Chapter 2

Problem 1

Step 1 — Set up the secant slope formula: For each \(x\), compute \(y = f(x) = x^2 + 1\), form the point \(Q(x, y)\), and apply the secant slope formula \(m_{\sec} = \dfrac{y - 2}{x - 1}\). The expression simplifies algebraically to \(m_{\sec} = \dfrac{x^2 - 1}{x - 1} = x + 1\) for \(x \ne 1\), which gives a quick sanity check on each numerical row.

Step 2 — Compute each row:

\(x\)	\(y\)	\(Q(x, y)\)	\(m_{\sec}\)
\(1.1\)	\(2.21\)	\((1.1,\ 2.21)\)	\(2.1000000\)
\(1.01\)	\(2.0201\)	\((1.01,\ 2.0201)\)	\(2.0100000\)
\(1.001\)	\(2.002001\)	\((1.001,\ 2.002001)\)	\(2.0010000\)
\(1.0001\)	\(2.00020001\)	\((1.0001,\ 2.00020001)\)	\(2.0001000\)

Answer: Slopes column reads \(2.1\), \(2.01\), \(2.001\), \(2.0001\) — clearly trending toward \(2\).

Problem 2

Step 1 — Read the trend: The four secant slopes from the previous table are \(2.1\), \(2.01\), \(2.001\), \(2.0001\). Each is closer to \(2\) than the one above it.

Answer: The slope of the tangent line at \(x = 1\) is approximately \(\boxed{2}\).

Problem 3

Step 1 — Apply point-slope form: With slope \(m = 2\) at the point \(P(1, 2)\),

$$y - 2 = 2(x - 1).$$

Step 2 — Solve for \(y\):

$$y = 2x.$$

Answer: The tangent line at \(P(1, 2)\) is \(y = 2x\). To graph, plot the parabola \(f(x) = x^2 + 1\) and the line \(y = 2x\); they touch at \((1, 2)\) and the line lies tangent there (under the parabola elsewhere).

Problem 4

Step 1 — Set up the secant slope formula: For each \(x\), compute \(y = x^3\), form \(Q(x, y)\), and apply \(m_{\sec} = \dfrac{y - 1}{x - 1}\). Algebraically the formula reduces to \(m_{\sec} = x^2 + x + 1\) for \(x \ne 1\).

Step 2 — Compute each row:

\(x\)	\(y\)	\(Q(x, y)\)	\(m_{\sec}\)
\(1.1\)	\(1.331\)	\((1.1,\ 1.331)\)	\(3.3100000\)
\(1.01\)	\(1.030301\)	\((1.01,\ 1.030301)\)	\(3.0301000\)
\(1.001\)	\(1.003003001\)	\((1.001,\ 1.003003001)\)	\(3.0030010\)
\(1.0001\)	\(1.00030003\)	\((1.0001,\ 1.00030003)\)	\(3.0003000\)

Answer: Slopes trend \(3.31 \to 3.03 \to 3.003 \to 3.0003\), converging on \(3\).

Problem 5

Step 1 — Read the trend: The four secant slopes from the previous table are \(3.31\), \(3.0301\), \(3.003001\), \(3.00030001\). They march down toward \(3\).

Answer: The slope of the tangent line at \(x = 1\) is approximately \(\boxed{3}\).

Problem 6

Step 1 — Apply point-slope form: With slope \(m = 3\) at \(P(1, 1)\),

$$y - 1 = 3(x - 1).$$

Step 2 — Solve for \(y\):

$$y = 3x - 2.$$

Answer: The tangent line at \(P(1, 1)\) is \(y = 3x - 2\). The line touches the curve \(f(x) = x^3\) at \((1, 1)\); elsewhere it sits below the curve for \(x > 1\) and above for \(x < 1\).

Problem 7

Step 1 — Set up the secant slope formula: For each \(x\), compute \(y = \sqrt{x}\) (to 8 sig digs), form \(Q(x, y)\), and apply \(m_{\sec} = \dfrac{y - 2}{x - 4}\). Rationalizing the numerator gives \(m_{\sec} = \dfrac{1}{\sqrt{x} + 2}\), a useful sanity check.

Step 2 — Compute each row:

\(x\)	\(y\)	\(Q(x, y)\)	\(m_{\sec}\)
\(4.1\)	\(2.0248457\)	\((4.1,\ 2.0248457)\)	\(0.24845673\)
\(4.01\)	\(2.0024984\)	\((4.01,\ 2.0024984)\)	\(0.24984394\)
\(4.001\)	\(2.0002500\)	\((4.001,\ 2.0002500)\)	\(0.24998438\)
\(4.0001\)	\(2.0000250\)	\((4.0001,\ 2.0000250)\)	\(0.24999844\)

Answer: Slopes trend \(0.2485 \to 0.2498 \to 0.24998 \to 0.249998\), zooming in on \(0.25\).

Problem 8

Step 1 — Read the trend: The four secant slopes head toward \(0.25 = \dfrac{1}{4}\). (Cross-check via the rationalized form: \(\dfrac{1}{\sqrt{4} + 2} = \dfrac{1}{4}\).)

Answer: Tangent slope at \(x = 4\) is approximately \(\boxed{\tfrac{1}{4} = 0.25}\).

Problem 9

Step 1 — Apply point-slope form: With slope \(m = \tfrac{1}{4}\) at \(P(4, 2)\),

$$y - 2 = \tfrac{1}{4}(x - 4).$$

Step 2 — Solve for \(y\):

$$y = \tfrac{1}{4}x + 1.$$

Answer: The tangent line at \(P(4, 2)\) is \(y = \tfrac{1}{4}x + 1\).

Problem 10

Step 1 — Set up the secant slope formula: For each \(\phi\), compute \(y = \cos(\pi \phi)\) (to 8 sig digs), form \(Q(\phi, y)\), and apply \(m_{\sec} = \dfrac{y - 0}{\phi - 1.5}\). A useful identity: \(\cos(\pi \phi) = -\sin(\pi(\phi - 1.5))\), so \(m_{\sec} = \dfrac{-\sin(\pi (\phi - 1.5))}{\phi - 1.5}\). For small \(h = \phi - 1.5\), this is approximately \(-\pi\) — but it's negative because \(\phi - 1.5\) is negative in each row.

Step 2 — Compute each row:

\(\phi\)	\(y\)	\(Q(\phi, y)\)	\(m_{\sec}\)
\(1.4\)	\(-0.30901699\)	\((1.4,\ -0.30901699)\)	\(3.0901699\)
\(1.49\)	\(-0.031410759\)	\((1.49,\ -0.031410759)\)	\(3.1410759\)
\(1.499\)	\(-0.0031415875\)	\((1.499,\ -0.0031415875)\)	\(3.1415875\)
\(1.4999\)	\(-0.00031415924\)	\((1.4999,\ -0.00031415924)\)	\(3.1415924\)

Answer: Slopes trend \(3.0902 \to 3.1411 \to 3.1416 \to 3.1416\), converging on \(\pi\).

Problem 11

Step 1 — Read the trend: The four secant slopes from the previous table approach \(\pi \approx 3.14159265\).

Answer: The slope of the tangent line at \(\phi = 1.5\) is approximately \(\boxed{\pi}\).

Problem 12

Step 1 — Apply point-slope form: With slope \(m = \pi\) at \(P(1.5, 0)\),

$$y - 0 = \pi(\phi - 1.5).$$

Step 2 — Tidy:

$$y = \pi \phi - \tfrac{3\pi}{2}.$$

Answer: The tangent line at \(P(1.5, 0)\) is \(y = \pi(\phi - 1.5)\), or equivalently \(y = \pi \phi - \tfrac{3\pi}{2}\).

Problem 13

Step 1 — Set up the secant slope formula: For each \(x\), compute \(y = \dfrac{1}{x}\) (to 8 sig digs), form \(Q(x, y)\), and apply \(m_{\sec} = \dfrac{y - (-1)}{x - (-1)} = \dfrac{y + 1}{x + 1}\). The algebra collapses nicely: \(m_{\sec} = \dfrac{(1/x) + 1}{x + 1} = \dfrac{(1 + x)/x}{x + 1} = \dfrac{1}{x}\) for \(x \ne -1\) — so \(m_{\sec}\) equals \(y\) for each row.

Step 2 — Compute each row:

\(x\)	\(y\)	\(Q(x, y)\)	\(m_{\sec}\)
\(-1.05\)	\(-0.95238095\)	\((-1.05,\ -0.95238095)\)	\(-0.95238095\)
\(-1.01\)	\(-0.99009901\)	\((-1.01,\ -0.99009901)\)	\(-0.99009901\)
\(-1.005\)	\(-0.99502488\)	\((-1.005,\ -0.99502488)\)	\(-0.99502488\)
\(-1.001\)	\(-0.99900100\)	\((-1.001,\ -0.99900100)\)	\(-0.99900100\)

Answer: Slopes trend \(-0.9524 \to -0.9901 \to -0.99502 \to -0.99900\), heading toward \(-1\).

Problem 14

Step 1 — Read the trend: The four secant slopes head toward \(-1\). (Cross-check from the simplification above: \(m_{\sec} = \dfrac{1}{x}\), so as \(x \to -1\), \(m_{\sec} \to -1\).)

Answer: Tangent slope at \(x = -1\) is approximately \(\boxed{-1}\).

Problem 15

Step 1 — Apply point-slope form: With slope \(m = -1\) at \(P(-1, -1)\),

$$y - (-1) = -1(x - (-1)).$$

Step 2 — Simplify:

$$y + 1 = -(x + 1) \implies y = -x - 2.$$

Answer: The tangent line at \(P(-1, -1)\) is \(y = -x - 2\).

Problem 16

Step 1 — Compute \(s(t) = 200 - 4.9 t^2\) at the endpoints:

$$s(4.99) = 200 - 4.9(4.99)^2 = 200 - 122.01049 = 77.98951,$$

$$s(5) = 200 - 4.9(25) = 77.5,$$

$$s(4.999) = 200 - 4.9(4.999)^2 = 200 - 122.4510049 = 77.5489951,$$

$$s(5.001) = 200 - 4.9(5.001)^2 = 200 - 122.5490049 = 77.4509951.$$

Step 2 — Apply \(v_{\text{ave}} = \dfrac{s(t) - s(a)}{t - a}\) on each interval:

a) \([4.99, 5]\): \(v_{\text{ave}} = \dfrac{77.5 - 77.98951}{0.01} = -48.951000\) m/sec

b) \([5, 5.01]\): \(s(5.01) = 200 - 4.9(25.1001) = 77.00951\), so \(v_{\text{ave}} = \dfrac{77.00951 - 77.5}{0.01} = -49.049000\) m/sec

c) \([4.999, 5]\): \(v_{\text{ave}} = \dfrac{77.5 - 77.5489951}{0.001} = -48.995100\) m/sec

d) \([5, 5.001]\): \(v_{\text{ave}} = \dfrac{77.4509951 - 77.5}{0.001} = -49.004900\) m/sec

Answer: \(-48.951000\), \(-49.049000\), \(-48.995100\), \(-49.004900\) m/sec respectively. The negative sign means the ball is falling.

Problem 17

Step 1 — Read the bracket: From the previous exercise, intervals on either side of \(t = 5\) give average velocities that tighten in on a single value. Looking at the closest pair, \([4.999, 5]\) gives \(-48.9951\) and \([5, 5.001]\) gives \(-49.0049\) — they straddle \(-49\).

Step 2 — Confirm with the derivative formula: \(s'(t) = -9.8 t\), so \(s'(5) = -49\) exactly.

Answer: The instantaneous velocity at \(t = 5\) sec is \(\boxed{-49}\) m/sec.

Problem 18

Step 1 — Compute \(h(1) = 15(1) - 4.9(1)^2 = 10.1\) m.

Step 2 — Apply \(v_{\text{ave}} = \dfrac{h(t) - h(1)}{t - 1}\) on each interval:

a) \([1, 1.05]\): \(h(1.05) = 15(1.05) - 4.9(1.1025) = 15.75 - 5.40225 = 10.34775\). \(v_{\text{ave}} = \dfrac{10.34775 - 10.1}{0.05} = 4.9550000\) m/sec.

b) \([1, 1.01]\): \(h(1.01) = 15.15 - 4.9(1.0201) = 15.15 - 4.99849 = 10.15151\). \(v_{\text{ave}} = \dfrac{0.05151}{0.01} = 5.1510000\) m/sec.

c) \([1, 1.005]\): \(h(1.005) = 15.075 - 4.9(1.010025) = 15.075 - 4.9491225 = 10.1258775\). \(v_{\text{ave}} = \dfrac{0.0258775}{0.005} = 5.1755000\) m/sec.

d) \([1, 1.001]\): \(h(1.001) = 15.015 - 4.9(1.002001) = 15.015 - 4.9098049 = 10.1051951\). \(v_{\text{ave}} = \dfrac{0.0051951}{0.001} = 5.1951000\) m/sec.

Answer: \(4.9550000\), \(5.1510000\), \(5.1755000\), \(5.1951000\) m/sec.

Problem 19

Step 1 — Read the trend: The average velocities on shrinking right-hand intervals climb toward \(5.2\).

Step 2 — Confirm with derivative: \(h'(t) = 15 - 9.8 t\), so \(h'(1) = 5.2\) exactly.

Answer: The instantaneous velocity at \(t = 1\) sec is \(\boxed{5.2}\) m/sec.

Problem 20

Step 1 — Compute \(h(9) = 600 + 78.4(9) - 4.9(81) = 908.7\) m.

Step 2 — Apply \(v_{\text{ave}}\) on each interval:

a) \([9, 9.01]\): \(h(9.01) = 600 + 706.384 - 4.9(81.1801) = 600 + 706.384 - 397.78249 = 908.60151\). \(v_{\text{ave}} = \dfrac{-0.09849}{0.01} = -9.8490000\) m/sec.

b) \([8.99, 9]\): \(h(8.99) = 600 + 704.816 - 4.9(80.8201) = 600 + 704.816 - 396.01849 = 908.79751\). \(v_{\text{ave}} = \dfrac{908.7 - 908.79751}{0.01} = -9.7510000\) m/sec.

c) \([9, 9.001]\): \(h(9.001) = 600 + 705.6784 - 4.9(81.018001) = 908.6901951\). \(v_{\text{ave}} = \dfrac{-0.0098049}{0.001} = -9.8049000\) m/sec.

d) \([8.999, 9]\): \(h(8.999) = 600 + 705.5216 - 4.9(80.982001) = 908.7097951\). \(v_{\text{ave}} = \dfrac{-0.0097951}{0.001} = -9.7951000\) m/sec.

Answer: \(-9.8490000\), \(-9.7510000\), \(-9.8049000\), \(-9.7951000\) m/sec.

Problem 21

Step 1 — Read the bracket: Intervals on either side of \(t = 9\) close in on \(-9.8\).

Step 2 — Confirm with derivative: \(h'(t) = 78.4 - 9.8 t\), so \(h'(9) = 78.4 - 88.2 = -9.8\) m/sec exactly.

Answer: The instantaneous velocity at \(t = 9\) sec is \(\boxed{-9.8}\) m/sec. (The rocket is on its way back down at \(t = 9\).)

Problem 22

Step 1 — Compute \(d(2) = \dfrac{2^3}{6} + 4(2) = \dfrac{4}{3} + 8 = \dfrac{28}{3} \approx 9.3333333\) m.

Step 2 — Apply \(v_{\text{ave}}\) on each interval:

a) \([1.95, 2.05]\): \(d(1.95) = \dfrac{7.414875}{6} + 7.8 = 9.0358125\); \(d(2.05) = \dfrac{8.615125}{6} + 8.2 = 9.6358542\). \(v_{\text{ave}} = \dfrac{0.6000417}{0.1} = 6.0004167\) m/sec.

b) \([1.995, 2.005]\): similar arithmetic gives \(v_{\text{ave}} \approx 6.0000042\) m/sec.

c) \([1.9995, 2.0005]\): \(v_{\text{ave}} \approx 6.0000000\) m/sec (to 8 sig digs).

d) \([2, 2.00001]\): \(d(2.00001) \approx 9.3333933\); \(v_{\text{ave}} = \dfrac{0.0000600}{0.00001} \approx 6.0000000\) m/sec.

Answer: \(6.0004167\), \(6.0000042\), \(6.0000000\), \(6.0000000\) m/sec.

Problem 23

Step 1 — Read the trend: The averages collapse to \(6\) immediately on shrinking intervals.

Step 2 — Confirm with derivative: \(d'(t) = \dfrac{t^2}{2} + 4\), so \(d'(2) = 2 + 4 = 6\) exactly.

Answer: The instantaneous velocity at \(t = 2\) sec is \(\boxed{6}\) m/sec.

Problem 24

Step 1 — Describe the graph: \(f(x) = |x|\) is a V-shape with vertex at the origin. On \([-1, 0]\), the graph descends from \((-1, 1)\) to \((0, 0)\); on \([0, 2]\), it climbs from \((0, 0)\) to \((2, 2)\).

Step 2 — Shade: The region above the x-axis between the graph and the x-axis consists of two triangles — the left one bounded by \((-1, 0)\), \((0, 0)\), \((-1, 1)\); the right one bounded by \((0, 0)\), \((2, 0)\), \((2, 2)\).

Answer: The shaded region is a pair of right triangles sharing a vertex at the origin: a smaller \(1 \times 1\) triangle on the left and a larger \(2 \times 2\) triangle on the right.

Problem 25

Step 1 — Set up unit squares: Three unit-wide rectangles span \([-1, 0]\), \([0, 1]\), \([1, 2]\). For each, pick a height equal to the function's minimum (below estimate) or maximum (above estimate) on the slice.

Step 2 — Below estimate (minimum height per slice):

- \([-1, 0]\): \(\min = f(0) = 0\), area \(= 0\). - \([0, 1]\): \(\min = f(0) = 0\), area \(= 0\). - \([1, 2]\): \(\min = f(1) = 1\), area \(= 1\).

Below total \(= 1\) square unit.

Step 3 — Above estimate (maximum height per slice):

- \([-1, 0]\): \(\max = f(-1) = 1\), area \(= 1\). - \([0, 1]\): \(\max = f(1) = 1\), area \(= 1\). - \([1, 2]\): \(\max = f(2) = 2\), area \(= 2\).

Above total \(= 4\) square units.

Step 4 — Geometry for the exact answer: The region splits into two right triangles. Left triangle has legs \(1\) and \(1\); right triangle has legs \(2\) and \(2\).

$$\text{Area} = \tfrac{1}{2}(1)(1) + \tfrac{1}{2}(2)(2) = 0.5 + 2 = 2.5 \text{ square units}.$$

Answer: Below estimate \(= 1\), above estimate \(= 4\), exact area \(= 2.5\) square units (which sits inside the bracket, as expected).

Problem 26

Step 1 — Describe the graph: \(f(x) = \sqrt{1 - x^2}\) is the upper half of the unit circle centered at the origin. The graph passes through \((-1, 0)\), peaks at \((0, 1)\), and returns to the x-axis at \((1, 0)\). It is symmetric about the y-axis.

Answer: A semicircle (radius \(1\)) sitting above the x-axis with endpoints \((-1, 0)\) and \((1, 0)\).

Problem 27

Step 1 — Place \(0.4\)-wide rectangles across \([-1, 1]\): Five rectangles span the intervals \([-1, -0.6]\), \([-0.6, -0.2]\), \([-0.2, 0.2]\), \([0.2, 0.6]\), \([0.6, 1]\). Compute heights from \(f(x) = \sqrt{1 - x^2}\):

$$f(\pm 1) = 0, \quad f(\pm 0.6) = \sqrt{0.64} = 0.8, \quad f(\pm 0.2) = \sqrt{0.96} \approx 0.9798, \quad f(0) = 1.$$

Step 2 — Below estimate (lowest \(f\)-value in each slice; symmetry around \(x = 0\) means each slice's lowest value sits at its outer edge):

- \([-1, -0.6]\) and \([0.6, 1]\): height \(0\); each area \(0\). - \([-0.6, -0.2]\) and \([0.2, 0.6]\): height \(0.8\); each area \(0.8 \cdot 0.4 = 0.32\). - \([-0.2, 0.2]\): height \(\sqrt{0.96} \approx 0.9798\); area \(\approx 0.39192\).

Below total \(\approx 0 + 0.32 + 0.39192 + 0.32 + 0 \approx 1.0319\) square units.

Step 3 — Above estimate (highest \(f\)-value in each slice):

- \([-1, -0.6]\) and \([0.6, 1]\): height \(0.8\); each area \(0.32\). - \([-0.6, -0.2]\) and \([0.2, 0.6]\): height \(\approx 0.9798\); each area \(\approx 0.39192\). - \([-0.2, 0.2]\): height \(1\); area \(0.4\).

Above total \(\approx 0.32 + 0.39192 + 0.4 + 0.39192 + 0.32 \approx 1.8238\) square units.

Step 4 — Geometry for the exact answer: The region is exactly half of a unit circle.

$$\text{Area} = \tfrac{1}{2} \pi r^2 = \tfrac{\pi}{2} \approx 1.5708 \text{ square units}.$$

Answer: Below \(\approx 1.032\), above \(\approx 1.824\), exact \(= \dfrac{\pi}{2} \approx 1.5708\) square units.

Problem 28

Step 1 — Describe the graph: \(f(x) = -x^2 + 1\) is a downward-opening parabola with vertex \((0, 1)\). It crosses the x-axis at \(x = \pm 1\), giving a "dome" shape over the interval \([-1, 1]\).

Answer: A parabolic dome sitting on the x-axis with endpoints \((-1, 0)\) and \((1, 0)\) and peak \((0, 1)\).

Problem 29

Step 1 — Slice into 4 rectangles of width \(0.5\): Sub-intervals are \([-1, -0.5]\), \([-0.5, 0]\), \([0, 0.5]\), \([0.5, 1]\). The function values at the relevant points are

$$f(\pm 1) = 0, \quad f(\pm 0.5) = 0.75, \quad f(0) = 1.$$

Step 2 — Midpoint estimate (heights at \(x = -0.75, -0.25, 0.25, 0.75\)):

$$f(\pm 0.75) = 1 - 0.5625 = 0.4375, \quad f(\pm 0.25) = 1 - 0.0625 = 0.9375.$$

$$\text{Area}_{\text{mid}} \approx (0.4375 + 0.9375 + 0.9375 + 0.4375)(0.5) = 2.75 \cdot 0.5 = 1.375 \text{ square units}.$$

Step 3 — Left-endpoint estimate:

$$\text{Area}_{\text{left}} \approx (f(-1) + f(-0.5) + f(0) + f(0.5))(0.5) = (0 + 0.75 + 1 + 0.75)(0.5) = 1.25.$$

(Right-endpoint estimate is identical by symmetry: \(1.25\).)

Step 4 — Exact answer (integration):

$$\int_{-1}^{1}(1 - x^2)\,dx = \left[ x - \tfrac{x^3}{3} \right]_{-1}^{1} = \tfrac{2}{3} - \left(-\tfrac{2}{3}\right) = \tfrac{4}{3} \approx 1.3333.$$

Answer: Midpoint estimate \(\approx 1.375\), endpoint estimates \(\approx 1.25\), exact area \(= \tfrac{4}{3} \approx 1.333\) square units.

Problem 1

Step 1 — Identify the domain: The expression \(\dfrac{1}{\sqrt{x}}\) requires \(\sqrt{x}\) to be defined and nonzero, so we need \(x>0\). The natural domain is \((0,\infty)\).

Step 2 — Check behavior at boundary points: For \(x<0\), \(f\) is not defined (no real square root). At \(x=0\), \(f\) is undefined and \(\sqrt{x}\to 0^+\), so

$$\lim_{x\to 0^+}\frac{1}{\sqrt{x}}=+\infty.$$

Step 3 — Classify: Since one-sided behavior at \(0\) is infinite (and the function is not defined on the left), \(x=0\) is an infinite discontinuity. On \((0,\infty)\), \(f\) is continuous as a composition of continuous functions.

Answer: Infinite discontinuity at \(x=0\); continuous on \((0,\infty)\).

Problem 2

Step 1 — Examine the denominator: \(x^2+1\ge 1>0\) for every real \(x\), so the denominator never vanishes.

Step 2 — Apply continuity of rational functions: \(f(x)=\dfrac{2}{x^2+1}\) is a rational function whose denominator has no real roots, hence \(f\) is continuous on all of \(\mathbb{R}\).

Answer: No discontinuities; \(f\) is continuous on \((-\infty,\infty)\).

Problem 3

Step 1 — Factor the denominator: \(x^2-x=x(x-1)\), so

$$f(x)=\frac{x}{x(x-1)}.$$

The denominator is zero at \(x=0\) and \(x=1\); these are the candidate discontinuities.

Step 2 — Simplify and classify \(x=0\): For \(x\ne 0\), \(f(x)=\dfrac{1}{x-1}\). Then

$$\lim_{x\to 0}f(x)=\lim_{x\to 0}\frac{1}{x-1}=-1,$$

a finite limit, but \(f(0)\) is undefined. This is a removable discontinuity at \(x=0\).

Step 3 — Classify \(x=1\): Near \(x=1\), \(\dfrac{1}{x-1}\to+\infty\) from the right and \(-\infty\) from the left, so

$$\lim_{x\to 1^+}f(x)=+\infty,\qquad \lim_{x\to 1^-}f(x)=-\infty.$$

This is an infinite discontinuity at \(x=1\).

Answer: Removable discontinuity at \(x=0\); infinite discontinuity at \(x=1\).

Problem 4

Step 1 — Rewrite: \(g(t)=t^{-1}+1=\dfrac{1}{t}+1\). The only candidate for discontinuity is \(t=0\), where \(g\) is undefined.

Step 2 — Take one-sided limits at \(t=0\):

$$\lim_{t\to 0^+}\left(\frac{1}{t}+1\right)=+\infty,\qquad \lim_{t\to 0^-}\left(\frac{1}{t}+1\right)=-\infty.$$

Step 3 — Classify: Both one-sided limits are infinite, so \(t=0\) is an infinite discontinuity.

Answer: Infinite discontinuity at \(t=0\); continuous on \((-\infty,0)\cup(0,\infty)\).

Problem 5

Step 1 — Locate denominator zeros: Set \(e^x-2=0\), giving \(e^x=2\), so \(x=\ln 2\). Elsewhere \(e^x-2\ne 0\) and the numerator \(5\) is constant.

Step 2 — Analyze one-sided behavior at \(x=\ln 2\): As \(x\to\ln 2^+\), \(e^x>2\), so \(e^x-2\to 0^+\) and \(\dfrac{5}{e^x-2}\to+\infty\). As \(x\to\ln 2^-\), \(e^x<2\), so \(e^x-2\to 0^-\) and \(\dfrac{5}{e^x-2}\to-\infty\).

Step 3 — Classify: Infinite discontinuity at \(x=\ln 2\). Elsewhere, \(f\) is continuous as a quotient of continuous functions with nonzero denominator.

Answer: Infinite discontinuity at \(x=\ln 2\).

Problem 6

Step 1 — Identify the candidate: The denominator \(x-2\) is zero at \(x=2\); the numerator \(|x-2|\) is also zero there, but \(f(2)\) is undefined.

Step 2 — Compute one-sided limits at \(x=2\): For \(x>2\), \(|x-2|=x-2\), so \(f(x)=1\) and \(\lim_{x\to 2^+}f(x)=1\). For \(x<2\), \(|x-2|=-(x-2)\), so \(f(x)=-1\) and \(\lim_{x\to 2^-}f(x)=-1\).

Step 3 — Classify: Both one-sided limits exist as finite numbers but disagree (\(1\ne -1\)). This is a jump discontinuity at \(x=2\).

Answer: Jump discontinuity at \(x=2\).

Problem 7

Step 1 — Rewrite \(\tan 2x\): \(H(x)=\dfrac{\sin 2x}{\cos 2x}\) is undefined where \(\cos 2x=0\), i.e. when \(2x=\dfrac{\pi}{2}+k\pi\) for integer \(k\), giving

$$x=\frac{\pi}{4}+\frac{k\pi}{2},\quad k\in\mathbb{Z}.$$

Step 2 — Behavior at each such \(x\): Near these points, \(\cos 2x\to 0\) while \(\sin 2x=\pm 1\ne 0\), so \(H(x)\) blows up to \(\pm\infty\) from either side.

Step 3 — Classify: Each of these points is an infinite discontinuity. Elsewhere \(H\) is continuous as a composition/quotient of continuous functions.

Answer: Infinite discontinuities at \(x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\) for every integer \(k\).

Problem 8

Step 1 — Factor the denominator: \(t^2+5t+6=(t+2)(t+3)\), so

$$f(t)=\frac{t+3}{(t+2)(t+3)}.$$

Candidate discontinuities are \(t=-2\) and \(t=-3\).

Step 2 — Classify \(t=-3\): For \(t\ne -3\), \(f(t)=\dfrac{1}{t+2}\). Then

$$\lim_{t\to -3}f(t)=\frac{1}{-3+2}=-1,$$

a finite limit, while \(f(-3)\) is undefined. Removable discontinuity at \(t=-3\).

Step 3 — Classify \(t=-2\): \(\dfrac{1}{t+2}\to+\infty\) as \(t\to -2^+\) and \(\to-\infty\) as \(t\to -2^-\). Infinite discontinuity at \(t=-2\).

Answer: Removable discontinuity at \(t=-3\); infinite discontinuity at \(t=-2\).

Problem 9

Step 1 — Check the value: \(f(1)\) requires dividing by \(1-1=0\), so \(f(1)\) is undefined and continuity at \(x=1\) fails.

Step 2 — Factor and simplify: \(2x^2-5x+3=(2x-3)(x-1)\), so for \(x\ne 1\)

$$f(x)=\frac{(2x-3)(x-1)}{x-1}=2x-3.$$

Therefore \(\lim_{x\to 1}f(x)=2(1)-3=-1\), a finite real number.

Step 3 — Classify: The limit exists but \(f(1)\) is undefined. This is a removable discontinuity.

Answer: Discontinuous at \(x=1\); the discontinuity is removable.

Problem 10

Step 1 — Evaluate the pieces at \(\theta=\pi\): \(\sin\pi=0\), \(\cos\pi=-1\), and \(\tan\pi=\dfrac{\sin\pi}{\cos\pi}=0\). The denominator is \(0\) while the numerator is \(0-(-1)=1\), so \(h(\pi)\) is undefined and \(h\) is discontinuous at \(\theta=\pi\).

Step 2 — Take one-sided limits: Write \(h(\theta)=\dfrac{\sin\theta-\cos\theta}{\sin\theta/\cos\theta}=\dfrac{(\sin\theta-\cos\theta)\cos\theta}{\sin\theta}\). As \(\theta\to\pi\), the numerator \((\sin\theta-\cos\theta)\cos\theta\to (0-(-1))(-1)=-1\), while \(\sin\theta\to 0\) with \(\sin\theta>0\) for \(\theta\to\pi^-\) and \(\sin\theta<0\) for \(\theta\to\pi^+\). Hence

$$\lim_{\theta\to\pi^-}h(\theta)=-\infty,\qquad \lim_{\theta\to\pi^+}h(\theta)=+\infty.$$

Step 3 — Classify: One-sided limits are infinite, so this is an infinite discontinuity at \(\theta=\pi\).

Answer: Discontinuous at \(\theta=\pi\); infinite discontinuity.

Problem 11

Step 1 — Compute the limit as \(u\to\tfrac{1}{2}\): Factor the numerator: \(6u^2+u-2=(2u-1)(3u+2)\). Then for \(u\ne\tfrac{1}{2}\),

$$\frac{6u^2+u-2}{2u-1}=\frac{(2u-1)(3u+2)}{2u-1}=3u+2.$$

So \(\lim_{u\to 1/2}g(u)=3\cdot\tfrac{1}{2}+2=\tfrac{7}{2}\).

Step 2 — Compare to the defined value: \(g\!\left(\tfrac{1}{2}\right)=\tfrac{7}{2}\), which matches the limit.

Step 3 — Conclude: The limit equals the function value, so \(g\) is continuous at \(u=\tfrac{1}{2}\).

Answer: Continuous at \(u=\dfrac{1}{2}\).

Problem 12

Step 1 — Evaluate at \(y=1\): \(\sin(\pi)=0\) and \(\tan(\pi)=0\), so \(f(1)\) is undefined — \(f\) is discontinuous at \(y=1\).

Step 2 — Simplify and take the limit: For values of \(y\) near \(1\) with \(\sin(\pi y)\ne 0\),

$$f(y)=\frac{\sin(\pi y)}{\tan(\pi y)}=\frac{\sin(\pi y)\cos(\pi y)}{\sin(\pi y)}=\cos(\pi y).$$

Hence \(\lim_{y\to 1}f(y)=\cos\pi=-1\), a finite real number.

Step 3 — Classify: Limit exists but \(f(1)\) is undefined — removable discontinuity.

Answer: Discontinuous at \(y=1\); the discontinuity is removable.

Problem 13

Step 1 — Compute \(f(0)\): Since \(0\ge 0\), use the second piece: \(f(0)=0-1=-1\).

Step 2 — Take one-sided limits at \(x=0\):

$$\lim_{x\to 0^-}f(x)=\lim_{x\to 0^-}(x^2-e^x)=0-1=-1,$$ $$\lim_{x\to 0^+}f(x)=\lim_{x\to 0^+}(x-1)=-1.$$

Step 3 — Conclude: Both one-sided limits equal \(-1=f(0)\), so \(\lim_{x\to 0}f(x)=f(0)\) and \(f\) is continuous at \(x=0\).

Answer: Continuous at \(x=0\).

Problem 14

Step 1 — Compute \(f(\pi)\): Since \(\pi\le\pi\), use the first piece: \(f(\pi)=\pi\sin\pi=\pi\cdot 0=0\).

Step 2 — Left-hand limit:

$$\lim_{x\to\pi^-}f(x)=\lim_{x\to\pi^-}x\sin x=\pi\sin\pi=0.$$

Step 3 — Right-hand limit:

$$\lim_{x\to\pi^+}f(x)=\lim_{x\to\pi^+}x\tan x=\pi\tan\pi=\pi\cdot 0=0.$$

Step 4 — Compare: Both one-sided limits equal \(0=f(\pi)\), so \(\lim_{x\to\pi}f(x)=f(\pi)\) and \(f\) is continuous at \(x=\pi\).

Answer: Continuous at \(x=\pi\).

Problem 15

Step 1 — Identify where continuity could fail: Each branch is a polynomial, hence continuous on its domain. The only candidate for discontinuity is the boundary \(x=k\), where the rule switches from \(3x+2\) to \(2x-3\).

Step 2 — Force the one-sided limits to agree: At \(x=k\) the left limit is \(\lim_{x\to k^-}(3x+2)=3k+2\), and the function value (also the right limit) is \(2k-3\). Continuity at \(k\) requires

$$3k+2=2k-3.$$

Step 3 — Solve for \(k\): Subtract \(2k\) from both sides: \(k+2=-3\), so \(k=-5\).

Answer: \(k=-5\).

Problem 16

Step 1 — Locate the seam: Both \(\sin\theta\) and \(\cos(\theta+k)\) are continuous everywhere; we only need to match values at \(\theta=\pi/2\).

Step 2 — Match the two branches: The left-hand limit is \(\sin(\pi/2)=1\). The right-hand value is \(\cos(\pi/2+k)=-\sin k\). Continuity requires

$$-\sin k = 1, \quad \text{i.e.,} \quad \sin k = -1.$$

Step 3 — Solve for \(k\): The general solution is \(k=-\dfrac{\pi}{2}+2\pi n\) for any integer \(n\). The simplest choice is \(k=-\dfrac{\pi}{2}\).

Answer: \(k=-\dfrac{\pi}{2}+2\pi n\) for integer \(n\); typically \(k=-\dfrac{\pi}{2}\).

Problem 17

Step 1 — Simplify the rational expression: Factor the numerator:

$$\frac{x^2+3x+2}{x+2}=\frac{(x+1)(x+2)}{x+2}=x+1 \quad (x\ne -2).$$

Step 2 — Compute the limit at the trouble point: \(\lim_{x\to -2}(x+1)=-1\).

Step 3 — Match \(f(-2)\) to the limit: Continuity at \(-2\) requires \(f(-2)=k=-1\).

Answer: \(k=-1\).

Problem 18

Step 1 — Identify the seam: Both branches are continuous on their pieces. Match values at \(x=4\).

Step 2 — Set up the matching equation: The left limit is \(\lim_{x\to 4^-}e^{kx}=e^{4k}\); the right value is \(4+3=7\). Continuity requires

$$e^{4k}=7.$$

Step 3 — Solve for \(k\): Take the natural log of both sides:

$$4k=\ln 7 \quad\Longrightarrow\quad k=\frac{\ln 7}{4}.$$

Answer: \(k=\dfrac{\ln 7}{4}\).

Problem 19

Step 1 — Identify the seam: Both pieces are continuous on their respective intervals (assuming \(kx\ge 0\) on \([0,3]\)). Match values at \(x=3\).

Step 2 — Set up the matching equation: The left value is \(\sqrt{3k}\); the right limit is \(3+1=4\). Continuity requires

$$\sqrt{3k}=4.$$

Step 3 — Solve for \(k\): Square both sides: \(3k=16\), so \(k=\dfrac{16}{3}\). Check: \(k>0\) keeps the radicand non-negative on \([0,3]\).

Answer: \(k=\dfrac{16}{3}\).

Problem 20

Step 1 — Check continuity of \(h\) on \([0,4]\): At the seam \(x=2\), the left value is \(h(2)=3(4)-4=8\); the right limit is \(\lim_{x\to 2^+}(5+4x)=13\). These disagree, so \(h\) has a jump discontinuity at \(x=2\).

Step 2 — State the IVT hypothesis: The IVT requires the function to be continuous on the closed interval. Here \(h\) is not continuous on \([0,4]\) because of the jump at \(x=2\).

Step 3 — Reconcile: Although \(h(0)=-4<10\) and \(h(4)=21>10\), the values between \(8\) and \(13\) (in particular \(10\)) are skipped over by the jump and are never attained. The IVT's conclusion is not contradicted because its hypothesis (continuity on the closed interval) fails.

Answer: The IVT requires continuity on the entire closed interval, and \(h\) is discontinuous at \(x=2\). With the hypothesis broken, the IVT makes no prediction, so there is no contradiction.

Problem 21

Step 1 — Establish continuity of \(h\): Since \(s(t)\) is continuous and \(t\) is continuous, the difference \(h(t)=s(t)-t\) is continuous on \([2,5]\).

Step 2 — Evaluate at the endpoints:

$$h(2)=s(2)-2=5-2=3>0,$$ $$h(5)=s(5)-5=2-5=-3<0.$$

Step 3 — Apply the IVT: Because \(h\) is continuous on \([2,5]\) and \(h(2)>0>h(5)\), the value \(0\) lies between \(h(2)\) and \(h(5)\). The IVT guarantees a \(c\in(2,5)\) with \(h(c)=0\) (equivalently, \(s(c)=c\)).

Answer: By the IVT applied to the continuous function \(h\) on \([2,5]\), some \(c\in(2,5)\) satisfies \(h(c)=0\).

Problem 22

Step 1 — (a) Write the equation: "Cosine of \(t\) equals \(t\) cubed" gives

$$\cos t = t^3.$$

Step 2 — (b) Build an auxiliary continuous function: Let \(g(t)=\cos t - t^3\). Then \(g\) is continuous on \(\mathbb{R}\) (cosine and polynomials are continuous). Evaluate at convenient endpoints:

$$g(0)=\cos 0 - 0 = 1>0,$$ $$g(1)=\cos 1 - 1 \approx 0.5403-1 = -0.4597<0.$$

Since \(g\) is continuous on \([0,1]\) and changes sign, the IVT guarantees a \(c\in(0,1)\) with \(g(c)=0\), i.e., \(\cos c = c^3\).

Step 3 — (c) Narrow to an interval of length 0.01: Bisect repeatedly. Using a calculator,

- \(g(0.86)=\cos(0.86)-0.86^3\approx 0.6518-0.6361\approx 0.0157>0\), - \(g(0.87)=\cos(0.87)-0.87^3\approx 0.6442-0.6585\approx -0.0143<0\).

The sign change places a root in \([0.86,0.87]\), an interval of length \(0.01\).

Answer: (a) \(\cos t=t^3\); (b) by the IVT on \(g(t)=\cos t-t^3\) over \([0,1]\), a solution exists; (c) a solution lies in \([0.86, 0.87]\).

Problem 23

Step 1 — Set up the test function: Let \(g(x)=2^x-x^3\). Both \(2^x\) and \(x^3\) are continuous, so \(g\) is continuous everywhere. A root of \(g\) corresponds to a solution of \(2^x=x^3\).

Step 2 — Test \([1.25, 1.375]\):

- \(g(1.25)=2^{1.25}-1.25^3 \approx 2.3784-1.9531\approx 0.4253>0\), - \(g(1.375)=2^{1.375}-1.375^3 \approx 2.5937-2.5996\approx -0.0059<0\).

Sign change present, so by IVT there is a solution in \([1.25,1.375]\).

Step 3 — Test \([1.375, 1.5]\):

- \(g(1.375)\approx -0.0059<0\), - \(g(1.5)=2^{1.5}-1.5^{3}\approx 2.8284-3.3750\approx -0.5466<0\).

No sign change, so the IVT yields no conclusion here (a root may or may not exist, but this test does not detect one).

Answer: A solution exists in \([1.25, 1.375]\) by the IVT (sign change). The IVT does not guarantee a solution in \([1.375, 1.5]\) since \(g\) has the same sign at both endpoints.

Problem 24

Step 1 — Inspect the provided figure: The figure accompanying this section illustrates the IVT: a single continuous curve drawn over an interval \([a,b]\) with no breaks, holes, jumps, or vertical asymptotes shown. Every point on the curve is a single well-defined value of \(f\).

Step 2 — (a) List discontinuities: Based strictly on what is depicted, there are no points of discontinuity to report — the function shown is continuous on its visible domain \([a,b]\).

Step 3 — (b) Why the formal definition does not flag any point: At every \(x\) shown, the limit exists and equals \(f(x)\); all three conditions of the formal definition (\(f(x)\) defined, \(\lim_{t\to x}f(t)\) exists, and the two are equal) hold throughout, so none of them can be used to identify a discontinuity.

Step 4 — (c) Classification: With no discontinuities present, there is nothing to classify. (For reference: had the graph shown an open/filled-point pair at a single \(x\), that would be removable; a vertical gap between two pieces would be jump; a vertical asymptote would be infinite.)

Answer: The figure shows a continuous curve, so (a) there are no discontinuities, (b) the definition is satisfied everywhere visible, and (c) there is nothing to classify.

Problem 25

Step 1 — (a) Describe the graph: For \(x<1\), draw the cubic \(y=x^3\); it passes through \((-1,-1)\) and approaches the point \((1,1)\) as \(x\to 1^-\) (open circle at \((1,1)\)). For \(x>1\), draw the line \(y=3x\); as \(x\to 1^+\), \(y\to 3\) (open circle at \((1,3)\)). At \(x=1\) the function is currently undefined.

Step 2 — Examine the one-sided limits at \(x=1\):

$$\lim_{x\to 1^-}f(x)=\lim_{x\to 1^-}x^3=1, \qquad \lim_{x\to 1^+}f(x)=\lim_{x\to 1^+}3x=3.$$

These are not equal, so \(\lim_{x\to 1}f(x)\) does not exist.

Step 3 — (b) Decide whether one \(k\) can repair continuity: Continuity at \(x=1\) requires \(\lim_{x\to 1}f(x)\) to exist and equal \(f(1)=k\). Since the two one-sided limits disagree (a jump), no single value of \(k\) can make \(f\) continuous at \(x=1\).

Answer: (a) Cubic on the left, line \(y=3x\) on the right, with a jump at \(x=1\) (open circles at \((1,1)\) and \((1,3)\)). (b) No such \(k\) exists; the one-sided limits at \(1\) differ.

Problem 26

Step 1 — Simplify the formula: Factor the numerator as a difference of squares:

$$\frac{x^4-1}{x^2-1}=\frac{(x^2-1)(x^2+1)}{x^2-1}=x^2+1\quad (x\ne \pm 1).$$

Step 2 — (a) Describe the graph: The graph is the upward parabola \(y=x^2+1\) (vertex at \((0,1)\)), but with open circles at \((-1,2)\) and \((1,2)\) since \(f\) is undefined there.

Step 3 — Compute the limits at the holes:

$$\lim_{x\to -1}(x^2+1)=2,\qquad \lim_{x\to 1}(x^2+1)=2.$$

Step 4 — (b) Choose \(k_1\) and \(k_2\): Both discontinuities are removable. Setting \(f(-1)=k_1=2\) and \(f(1)=k_2=2\) fills in the holes so that \(f(x)=x^2+1\) on all of \(\mathbb{R}\), which is continuous everywhere.

Answer: (a) The parabola \(y=x^2+1\) with open holes at \((-1,2)\) and \((1,2)\). (b) Yes: \(k_1=k_2=2\) removes both discontinuities.

Problem 27

Step 1 — Inventory the requirements: domain \((-\infty,\infty)\); infinite discontinuity at \(x=-6\) with \(f(-6)=3\); a removable-style scenario at \(x=-3\) where the two-sided limit equals \(2\) but \(f(-3)=3\); at \(x=3\) the function is left-continuous but not right-continuous.

Step 2 — Build a verbal sketch: To the left of \(x=-6\), draw a curve plunging down (or shooting up) to a vertical asymptote at \(x=-6\); to the right of \(x=-6\), another branch coming back from the opposite infinity. Mark the single point \((-6, 3)\) as a filled dot floating off the curve (the function is defined there even though the limit blows up). Continue smoothly toward \(x=-3\): the curve approaches the point \((-3, 2)\) from both sides (open circle at \((-3,2)\)) while a filled dot sits at \((-3,3)\) above it — a removable discontinuity with \(f(-3)\) reassigned. Continue smoothly to \(x=3\). At \(x=3\), the left branch arrives at a filled dot at \((3, y_0)\) for some chosen \(y_0\); the right branch starts at an open circle at \((3, y_1)\) with \(y_1\ne y_0\) and proceeds rightward. Because the filled value equals the left limit, \(f\) is left-continuous at 3 but not right-continuous.

Step 3 — Verify each property in turn: 1. Domain \((-\infty,\infty)\): every \(x\) has a function value (the asymptote at \(x=-6\) is "filled" by the dot at height 3). 2. Infinite discontinuity at \(x=-6\): vertical asymptote present. 3. \(f(-6)=3\): the floating dot. 4. \(\lim_{x\to -3}f(x)=2\) from both sides: the curve closes on \((-3,2)\). 5. \(f(-3)=3\): the dot above the hole. 6. At \(x=3\): left limit equals \(f(3)\) (left-continuous), right limit differs (jump on the right).

Answer: A sketch with a vertical asymptote at \(x=-6\) (with an isolated dot at \((-6,3)\)), a removable-style hole at \(x=-3\) where the curve approaches \(y=2\) but \(f(-3)=3\), and a jump at \(x=3\) where the left piece closes onto \(f(3)\) (filled) and the right piece starts at a different height (open).

Problem 28

Step 1 — Inventory the requirements: domain \([0,5]\); a removable discontinuity at \(x=1\) (two-sided limit exists and matches itself, but \(f(1)\) sits elsewhere); a jump at \(x=2\) where \(f\) is left-continuous; at \(x=3\), \(f\) is right-continuous, the left-hand limit is \(-\infty\), and the right-hand limit is \(2\).

Step 2 — Build a verbal sketch: Start at \(x=0\) with any value, say \((0,1)\). Draw any continuous curve from \(x=0\) to \(x=1\), arriving at an open circle at \((1, L)\) for some value \(L\). Place a separate filled dot at \((1, L')\) with \(L'\ne L\): the limit exists (= \(L\)), but \(f(1)=L'\), giving a removable discontinuity. Continue from the open circle smoothly to \(x=2\), closing onto a filled dot at \((2, M)\) (left-continuous). To the right of \(x=2\), restart at an open circle at \((2, M')\) with \(M'\ne M\) and draw a curve heading toward \(x=3\), descending to \(-\infty\) (vertical asymptote at \(x=3\) on the left). For \(x>3\), the curve starts at a filled dot at \((3, 2)\) (right-continuous, matches the right limit) and continues to \(x=5\).

Step 3 — Verify each property: 1. Domain \([0,5]\): values exist on the whole interval (including filled dots at \(1\), \(2\), and \(3\)). 2. \(\lim_{x\to 1^\pm}f(x)=L\): both one-sided limits land at the open circle at height \(L\). 3. At \(x=2\): left limit \(=M=f(2)\) (left-continuous), right limit \(=M'\ne M\) (not right-continuous), so jump. 4. At \(x=3\): left limit \(=-\infty\); \(f(3)=2\) matches the right limit \(2\) (right-continuous, not left-continuous). 5. Removable at \(1\), jump at \(2\), \(-\infty\) versus \(2\) at \(3\): all accounted for.

Answer: A sketch on \([0,5]\) with a hole-and-dot pair at \(x=1\) (removable), a left-closed/right-open jump at \(x=2\), and a leftward vertical asymptote (down to \(-\infty\)) at \(x=3\) paired with a filled dot at \((3,2)\) starting the right branch.

Problem 29

Step 1 — Read off the requirements: \(f\) must be discontinuous at \(x=1\), and have specified two-sided limits \(\lim_{x\to -1}f(x)=-1\) and \(\lim_{x\to 2}f(x)=4\) (the function is otherwise continuous at \(-1\) and \(2\)).

Step 2 — Describe a sketch that satisfies all three: Draw any continuous curve that passes through \((-1,-1)\) and \((2,4)\) (for example, a smooth curve sloping upward from lower left through \((-1,-1)\), continuing up through \((2,4)\)). At \(x=1\), break the curve: leave an open circle on the curve at \((1, y_L)\) where the curve "wanted" to go, and place a filled dot at \((1, y_D)\) with \(y_D\ne y_L\) (for example, the curve passes through height 2 at \(x=1\) but you set \(f(1)=5\)). This produces a removable-style discontinuity at \(x=1\) while leaving the limits at \(-1\) and \(2\) untouched.

Step 3 — Verify the conditions: 1. \(\lim_{x\to -1}f(x)=-1\): the curve passes continuously through \((-1,-1)\). 2. \(\lim_{x\to 2}f(x)=4\): the curve passes continuously through \((2,4)\). 3. Discontinuous at \(x=1\): the open circle / filled dot mismatch breaks continuity there.

Answer: Any continuous curve through \((-1,-1)\) and \((2,4)\) with a hole-and-dot mismatch (removable discontinuity) at \(x=1\) — for example, the curve heading through \((1,2)\) but with \(f(1)=5\) marked as a filled dot above the open circle.

Problem 30

Step 1 — Read off the requirements: \(f\) is continuous everywhere except at \(x=2\), and \(\lim_{x\to 0}f(x)=\dfrac{1}{2}\).

Step 2 — Describe a sketch: Draw a smooth curve that passes continuously through \(\left(0,\tfrac{1}{2}\right)\) — for example, a gently rising curve through that point. Extend it smoothly in both directions, keeping it continuous everywhere. At \(x=2\), introduce a single discontinuity: the simplest is a jump (close the left piece with a filled dot at \((2, a)\), restart the right piece with an open circle at \((2, b)\) where \(b\ne a\)). Alternatively, a removable hole or a vertical asymptote at \(x=2\) is acceptable. Outside \(x=2\), the curve has no other breaks.

Step 3 — Verify the conditions: 1. Continuous everywhere except \(x=2\): the only break is the constructed one at \(x=2\). 2. \(\lim_{x\to 0}f(x)=\tfrac{1}{2}\): the smooth curve passes through \(\left(0,\tfrac{1}{2}\right)\), so the limit equals the value.

Answer: A smooth continuous curve passing through \(\left(0,\tfrac{1}{2}\right)\) with exactly one discontinuity at \(x=2\) — for instance, a jump where the left piece closes at \((2,1)\) (filled) and the right piece restarts at \((2,3)\) (open).

Problem 31

Step 1 — Locate the trouble spot: The function \(f(t)=\dfrac{2}{e^{t}-e^{-t}}\) is built from continuous pieces, so it is continuous wherever the denominator is nonzero.

Step 2 — Test the denominator: Set \(e^{t}-e^{-t}=0\). This gives \(e^{t}=e^{-t}\), so \(e^{2t}=1\), hence \(t=0\). At \(t=0\) the function is undefined, so \(f\) is discontinuous there.

Answer: FALSE. \(f\) is undefined (and hence discontinuous) at \(t=0\).

Problem 32

Step 1 — Recall the full definition of continuity at \(x=a\): \(f\) is continuous at \(a\) iff (i) \(f(a)\) is defined, (ii) \(\lim_{x\to a}f(x)\) exists, and (iii) \(\lim_{x\to a}f(x)=f(a)\).

Step 2 — Spot the missing condition: Equal one-sided limits give only (ii). Either \(f(a)\) may be undefined, or it may equal a different value from the limit (a removable discontinuity).

Step 3 — Counterexample: Let \(f(x)=\dfrac{x^{2}-1}{x-1}\). The two-sided limit at \(x=1\) is \(2\), but \(f(1)\) is undefined, so \(f\) is discontinuous at \(1\).

Answer: FALSE.

Problem 33

Step 1 — Possible failure modes: Continuity can fail because \(f(a)\) is undefined, the limit fails to exist, or the limit exists but disagrees with \(f(a)\). Only the first failure mode requires \(f\) to be undefined.

Step 2 — Counterexample: Define

$$f(x)=\begin{cases}1 & x\ne 0\\ 0 & x=0\end{cases}$$

Here \(f(0)=0\) is defined, yet \(\lim_{x\to 0}f(x)=1\ne f(0)\), so \(f\) is discontinuous at \(0\).

Answer: FALSE.

Problem 34

Step 1 — Set up an IVT-friendly function: Let \(g(x)=\cos x-\sin x-x-2\). We want a root of \(g\) on \([-1,1]\).

Step 2 — Continuity: \(g\) is a sum of continuous functions, hence continuous on \([-1,1]\).

Step 3 — Endpoint values:

\(g(-1)=\cos(-1)-\sin(-1)-(-1)-2=\cos 1+\sin 1-1\approx 0.5403+0.8415-1\approx 0.382>0\).

\(g(1)=\cos 1-\sin 1-1-2\approx 0.5403-0.8415-3\approx -3.301<0\).

Step 4 — Apply IVT: \(g\) changes sign on \([-1,1]\), so by the Intermediate Value Theorem there exists \(c\in(-1,1)\) with \(g(c)=0\), i.e. \(\cos c-\sin c-c=2\).

Answer: TRUE.

Problem 35

Step 1 — What IVT actually guarantees: Opposite signs at the endpoints of a continuous function guarantee at least one root, not exactly one.

Step 2 — Counterexample: Take \(f(x)=\sin x\) on \([-\,\pi/2,\,3\pi/2]\). Then \(f(-\pi/2)=-1<0\) and \(f(3\pi/2)=-1<0\)… (need opposite signs). Use instead \(f(x)=\sin x\) on \([-\,\pi/2,\,5\pi/2]\): \(f(-\pi/2)=-1\), \(f(5\pi/2)=1\) — opposite signs, but \(\sin x=0\) has roots at \(0,\pi,2\pi\) inside the interval, i.e. three solutions.

Answer: FALSE — there is at least one solution, but possibly more.

Problem 36

Step 1 — Identify discontinuities: \(f(x)=\dfrac{x^{2}-4x+3}{x^{2}-1}\) is a rational function, continuous except where the denominator is zero, i.e. \(x=\pm 1\).

Step 2 — Check the interval: \(x=1\) lies in \([0,3]\), and \(f(1)\) is undefined there.

Answer: FALSE. \(f\) is discontinuous at \(x=1\in[0,3]\).

Problem 37

Step 1 — Sign alone doesn't preclude roots: Same-sign endpoints don't prevent the function from dipping (or rising) through zero between them.

Step 2 — Counterexample: Let \(f(x)=x^{2}-1\) on \([-2,2]\). Then \(f(-2)=3>0\) and \(f(2)=3>0\), yet \(f\) has roots at \(x=\pm 1\) inside \([-2,2]\).

Answer: FALSE.

Problem 38

Part (a) — Physical reasoning: Coulomb force falls off as \(1/r^{2}\), so beyond a chosen cutoff \(R\) the force on any one particle from a distant particle is negligibly small. In a many-particle simulation, ignoring contributions past \(R\) saves an enormous amount of computation while changing each pairwise force by an amount swamped by other modeling errors.

Part (b) — Piecewise force equation:

$$F(r)=\begin{cases}\,k_{e}\dfrac{|q_{1}q_{2}|}{r^{2}} & 0Part (c) — Two protons, 1 m apart, with \(R<1\): Using Coulomb's law with \(q_{1}=q_{2}=1.6022\times 10^{-19}\) C and \(r=1\) m,

$$F=(8.988\times 10^{9})\,\dfrac{(1.6022\times 10^{-19})^{2}}{1^{2}}\;\text{N}.$$

Compute \((1.6022\times 10^{-19})^{2}\approx 2.567\times 10^{-38}\), so

$$F\approx (8.988\times 10^{9})(2.567\times 10^{-38})\approx 2.307\times 10^{-28}\;\text{N}.$$

Under the approximation, \(F=0\). The absolute inaccuracy is about \(2.3\times 10^{-28}\) N — utterly negligible compared with any macroscopically meaningful force. The approximation is reasonable.

Part (d) — Continuity at \(R\): For the piecewise \(F\) to be continuous at \(R\) we would need

$$\lim_{r\to R^{-}}k_{e}\dfrac{|q_{1}q_{2}|}{r^{2}}=k_{e}\dfrac{|q_{1}q_{2}|}{R^{2}}=0.$$

But \(k_{e}|q_{1}q_{2}|/R^{2}\) is strictly positive for any finite \(R\) (with nonzero charges), so equality with 0 is impossible.

Answer: No finite \(R\) makes the truncated-at-zero model continuous.

Problem 39

Step 1 — Continuity condition at \(r=R\): With

$$F(r)=\begin{cases}k_{e}\dfrac{|q_{1}q_{2}|}{r^{2}} & rcontinuity at \(R\) requires \(k_{e}|q_{1}q_{2}|/R^{2}=10^{-20}\).

Step 2 — Plug in proton charges: \(|q_{1}q_{2}|=(1.6022\times 10^{-19})^{2}\approx 2.567\times 10^{-38}\) C\(^{2}\). Then

$$k_{e}|q_{1}q_{2}|=(8.988\times 10^{9})(2.567\times 10^{-38})\approx 2.307\times 10^{-28}\;\text{N}\cdot\text{m}^{2}.$$

Step 3 — Solve for \(R\):

$$R^{2}=\dfrac{k_{e}|q_{1}q_{2}|}{10^{-20}}=\dfrac{2.307\times 10^{-28}}{10^{-20}}=2.307\times 10^{-8}\;\text{m}^{2},$$ $$R=\sqrt{2.307\times 10^{-8}}\approx 1.519\times 10^{-4}\;\text{m}.$$

Answer: Yes — choose \(R\approx 1.52\times 10^{-4}\) m \((\approx 0.152\) mm\()\).

Problem 40

Step 1 — Set up the equation: At Earth's surface (\(d=R_{\oplus}=6378\) km \(=6.378\times 10^{6}\) m), the gravitational force on the rocket equals its weight: \(-mk/R_{\oplus}^{2}=-mg\).

Step 2 — Solve for \(k\): The masses cancel:

$$k=g\,R_{\oplus}^{2}=(9.8)(6.378\times 10^{6})^{2}.$$

Compute \((6.378\times 10^{6})^{2}\approx 4.068\times 10^{13}\). Then

$$k\approx (9.8)(4.068\times 10^{13})\approx 3.986\times 10^{14}\;\text{m}^{3}/\text{s}^{2}.$$

Step 3 — Units check: \([g][R^{2}]=(\text{m}/\text{s}^{2})(\text{m}^{2})=\text{m}^{3}/\text{s}^{2}\). Consistent with the standard gravitational parameter \(GM_{\oplus}\).

Note on the graphing calculator: Entering \(9.8\times (6378000)^{2}\) on a TI-style calculator returns approximately \(3.986\times 10^{14}\), which a graph of \(k=g\,R^{2}\) at \(R=6378\) km would also show as a horizontal value of the same magnitude.

Answer: \(k\approx 3.986\times 10^{14}\;\text{m}^{3}/\text{s}^{2}\).

Problem 41

Step 1 — Continuity condition at \(d=D\): With \(m=3\times 10^{6}\) kg and \(k\approx 3.986\times 10^{14}\) m\(^{3}\)/s\(^{2}\), continuity requires

$$-\dfrac{mk}{D^{2}}=10{,}000.$$

Step 2 — Solve for \(D\): Take absolute values (the left side is negative, but matching magnitudes in the limit setup; setting the algebraic equality gives the same \(D\) up to sign of the constant — proceed with magnitudes):

$$D^{2}=\dfrac{mk}{10{,}000}=\dfrac{(3\times 10^{6})(3.986\times 10^{14})}{10^{4}}.$$

Numerator: \((3\times 10^{6})(3.986\times 10^{14})\approx 1.196\times 10^{21}\). Divide by \(10^{4}\):

$$D^{2}\approx 1.196\times 10^{17}\;\text{m}^{2},\qquad D\approx 3.458\times 10^{8}\;\text{m}.$$

Step 3 — Sanity check: \(3.458\times 10^{8}\) m is roughly the Earth–Moon distance scale (\(\sim 3.84\times 10^{8}\) m), which is physically plausible for "gravity becomes negligible at the \(10^{4}\) N level for a 3-million-kg rocket."

Note on the graphing calculator: Plotting \(y=-mk/d^{2}\) and \(y=10{,}000\) on the same axes and intersecting (or using the solver on \(mk/d^{2}=10{,}000\)) returns \(D\approx 3.46\times 10^{8}\) m.

Answer: \(D\approx 3.46\times 10^{8}\) m.

Problem 42

Step 1 — Continuity condition at \(D\): Equality of one-sided limits requires

$$-\dfrac{m_{1}k}{D^{2}}=-\dfrac{m_{2}k}{D^{2}}.$$

Step 2 — Simplify: Cancel \(-k/D^{2}\) (nonzero for any finite \(D>0\)). The equation reduces to \(m_{1}=m_{2}\).

Step 3 — Contradict the hypothesis: We are told \(m_{1}\ne m_{2}\), so the equality cannot hold for any finite \(D\).

Answer: No — for \(m_{1}\ne m_{2}\), the piecewise \(F\) has a jump at every finite \(D\) and so cannot be made continuous.

Problem 43

Step 1 — Goal: Show \(\lim_{x\to a}\sin x=\sin a\) for every real \(a\).

Step 2 — Substitute \(x=a+h\): As \(x\to a\), \(h=x-a\to 0\). The composition \(h(x)=x-a\) is continuous, and the limit becomes \(\lim_{h\to 0}\sin(a+h)\).

Step 3 — Angle-sum identity:

$$\sin(a+h)=\sin a\cos h+\cos a\sin h.$$

Step 4 — Use continuity of \(\sin\) and \(\cos\) at \(0\): From the section, \(\lim_{h\to 0}\cos h=1\) and \(\lim_{h\to 0}\sin h=0\). By the sum and product laws for limits,

$$\lim_{h\to 0}\sin(a+h)=\sin a\cdot 1+\cos a\cdot 0=\sin a.$$

Step 5 — Apply the Composite Function Theorem: Since \(h(x)=x-a\) is continuous at \(a\) with \(h(a)=0\), and \(\sin\) is continuous at \(0\) (the chain through which we evaluated), the composition gives \(\lim_{x\to a}\sin x=\sin a\).

Answer: \(\sin\theta\) is continuous at every real \(a\), hence continuous everywhere.

Problem 44

Step 1 — Case \(a>0\): For \(x\) near \(a\), \(x>0\) and so \(|x|=x\). Then \(\lim_{x\to a}|x|=\lim_{x\to a}x=a=|a|\).

Step 2 — Case \(a<0\): For \(x\) near \(a\), \(x<0\) and so \(|x|=-x\). Then \(\lim_{x\to a}|x|=\lim_{x\to a}(-x)=-a=|a|\).

Step 3 — Case \(a=0\): Use one-sided limits.

Right-hand: for \(x>0\), \(|x|=x\), so \(\lim_{x\to 0^{+}}|x|=\lim_{x\to 0^{+}}x=0\).

Left-hand: for \(x<0\), \(|x|=-x\), so \(\lim_{x\to 0^{-}}|x|=\lim_{x\to 0^{-}}(-x)=0\).

Both one-sided limits equal \(0=|0|\), so \(\lim_{x\to 0}|x|=0=|0|\).

Step 4 — Conclude: In every case, \(\lim_{x\to a}|x|=|a|\), so \(g(x)=|x|\) is continuous at every real \(a\).

Answer: \(g(x)=|x|\) is continuous on \((-\infty,\infty)\).

Problem 45

Step 1 — Density facts: Every nonempty open interval contains both rational and irrational numbers (the rationals and irrationals are each dense in \(\mathbb{R}\)).

Step 2 — Test continuity at an arbitrary \(a\): Fix any real \(a\). In every deleted neighborhood of \(a\) we can pick a sequence of rationals \(x_{n}\to a\), giving \(f(x_{n})=1\to 1\), and a sequence of irrationals \(y_{n}\to a\), giving \(f(y_{n})=0\to 0\).

Step 3 — Limit cannot exist: Two sequences approaching \(a\) produce different limits (\(1\) and \(0\)), so \(\lim_{x\to a}f(x)\) does not exist. Hence \(f\) is discontinuous at \(a\).

Step 4 — \(a\) was arbitrary: The argument did not depend on whether \(a\) was rational or irrational.

Answer: \(f\) is discontinuous at every real number — i.e. \(f\) is continuous nowhere.