# Active Calculus

This appendix contains answers to all non-WeBWorK exercises in the text. For WeBWorK exercises, please use the HTML version of the text for access to answers and solutions.

### 1Understanding the Derivative1.1How do we measure velocity?1.1.4Exercises

#### 1.1.4.6.

1. $$s(15)-s(0) \approx -98.75\text{.}$$
2. \begin{align*} AV_{[0,15]} &= \frac{s(15)-s(0)}{15-0} \approx -6.58\\ AV_{[0,2]} &= \frac{s(2)-s(0)}{2-0} \approx -47.63\\ AV_{[1,6]} &= \frac{s(6)-s(1)}{6-1} \approx -13.25\\ AV_{[8,10]} &= \frac{s(10)-s(8)}{10-8} \approx -7.35 \end{align*}
3. Most negative average velocity on $$[0,4]\text{;}$$ most positive average velocity on $$[4,8]\text{.}$$
4. $$\frac{21.31+22.25}{2} = 21.78$$ feet per second.
5. The average velocities are negative; the instantaneous velocity was positive. Downward motion corresponds to negative average velocity; upward motion to positive average velocity.

#### 1.1.4.7.

1. Sketch a plot where the diver’s height at time $$t$$ is on the vertical axis. For instance, $$h(2.45) = 0\text{.}$$
2. $$AV_{[2.45,7]} \approx \frac{-3.5-0}{7-2.45}=\frac{-3.5}{4.55}=-0.7692$$ m/sec. The average velocity is not the same on every time interval within $$[2.45,7]\text{.}$$
3. When the diver is going upward, her velocity is positive. When she is going downward, her velocity is negative. At the peak of her dive and when her feet touch the bottom of the pool.
4. It looks like when the position function is steep, the velocity function’s value is farther away from zero, and that whenever the height/position function is rising/increasing, the velocity function has a positive value. Similarly, whenever the position function is decreasing, the velocity is negative.

#### 1.1.4.8.

1. $$15 957$$ people.
2. In an average year the population grew by about $$798$$ people/year.
3. The slope of a secant line through the points $$(a,f(a))$$ and $$(b,f(b))\text{.}$$
4. $$AV_{[0,20]} \approx 798$$ people per year.
5. \begin{align*} AV_{[5,10]} & \approx 734.50\\ AV_{[5,9]} & \approx 733.06\\ AV_{[5,8]} & \approx 731.62\\ AV_{[5,7]} & \approx 730.19\\ AV_{[5,6]} & \approx 728.7535 \end{align*}

### 1.2The notion of limit1.2.4Exercises

#### 1.2.4.5.

1. All real numbers except $$x = \pm 2\text{.}$$
2.  $$x$$ $$f(x)$$ $$2.1$$ $$-8.41$$ $$2.01$$ $$-8.0401$$ $$2.001$$ $$-8.004001$$ $$1.999$$ $$-7.996001$$ $$1.99$$ $$-7.9601$$ $$1.9$$ $$-7.61$$
$$\lim_{x \to 2} f(x) = -8\text{.}$$
3. $$\lim_{x \to 2} \frac{16-x^4}{x^2-4} = -8\text{.}$$
4. False.
5. False.

#### 1.2.4.6.

1. All real numbers except $$x = -3\text{.}$$
2.  $$x$$ $$g(x)$$ $$-2.9$$ $$-1$$ $$-2.99$$ $$-1$$ $$-2.999$$ $$-1$$ $$-3.001$$ $$1$$ $$-3.01$$ $$-1$$ $$-3.1$$ $$-1$$
The limit does not exist.
3. If $$x \gt -3\text{,}$$
\begin{equation*} -\frac{|x+3|}{x+3} = -\frac{x+3}{x+3} = -1; \end{equation*}
if $$x \lt -3\text{,}$$ it follows that
\begin{equation*} -\frac{|x+3|}{x+3} = -\frac{-(x+3)}{x+3} = +1\text{.} \end{equation*}
Hence the limit does not exist.
4. False.
5. False.

#### 1.2.4.8.

1. \begin{equation*} AV_{[1,1+h]} = \frac{100\cos(0.75(1+h)) \cdot e^{-0.2(1+h)} - 100\cos(0.75) \cdot e^{-0.2}}{h} \end{equation*}
2. \begin{equation*} \lim_{h \to 0} AV_{1, 1+h} \approx -53.837\text{.} \end{equation*}
3. The instantaneous velocity of the bungee jumper at the moment $$t = 1$$ is approximately $$-53.837$$ ft/sec.

### 1.3The derivative of a function at a point1.3.3Exercises

#### 1.3.3.6.

1. $$AV_{[-3,-1]} \approx 1.15\text{;}$$ $$AV_{[0,2]} \approx -0.4\text{.}$$
2. $$f'(-3) \approx 3\text{;}$$ $$f'(0) \approx -\frac{1}{2}\text{.}$$

#### 1.3.3.7.

1. For instance, you could let $$f(-3) = 3$$ and have $$f$$ pass through the points $$(-3,3)\text{,}$$ $$(-1,-2)\text{,}$$ $$(0,-3)\text{,}$$ $$(1,-2)\text{,}$$ and $$(3,-1)$$ and draw the desired tangent lines accordingly.
2. For instance, you could draw a function $$g$$ that passes through the points $$(-2,3)\text{,}$$ $$(-1,2)\text{,}$$ $$(1,0)\text{,}$$ $$(2,0)\text{,}$$ and $$(3,3)$$ in such a way that the tangent line at $$(-1,2)$$ is horizontal and the tangent line at $$(2,0)$$ has slope $$1\text{.}$$

#### 1.3.3.8.

1. $$AV_{[0,7]}=\frac{0.1175}{7} \approx 0.01679$$ billion people per year; $$P'(7) \approx 0.1762$$ billion people per year; $$P'(7) \gt AV_{[0,7]}\text{.}$$
2. $$AV_{[19,29]} \approx 0.02234$$ billion people/year.
3. We will say that today’s date is July 1, 2015, which means that $$t = 22.5\text{;}$$
\begin{equation*} P'(22.5) = \lim_{h \to 0} \frac{115(1.014)^{22.5+h}-115(1.014)^{22.5}}{h}; \end{equation*}
$$P'(22.5) \approx 0.02186$$ billions of people per year.
4. $$y - 1.57236 = 0.02186(t-22.5)\text{.}$$

#### 1.3.3.9.

1. All three approaches show that $$f'(2) = 1\text{.}$$
2. All three approaches show that $$f'(1) = -1\text{.}$$
3. All three approaches show that $$f'(1) = \frac{1}{2}\text{.}$$
4. All three approaches show that $$f'(1)$$ does not exist.
5. The first two approaches show that $$f'(\frac{\pi}{2}) = 0\text{.}$$

### 1.4The derivative function1.4.3Exercises

#### 1.4.3.6.

1. See the figure below.
2. See the figure below.
3. One example of a formula for $$f$$ is $$f(x) = \frac{1}{2}x^2 - 1\text{.}$$

#### 1.4.3.7.

1. $$g'(x) = 2x - 1\text{.}$$
2. $$p'(x) = 10x - 4\text{.}$$
3. The constants $$3$$ and $$12$$ don’t seem to affect the results at all. The coefficient $$-4$$ on the linear term in $$p(x)$$ appears to make the $$-4$$’’ appear in $$p'(x)= 10x - 4\text{.}$$ The leading coefficient $$5$$ in $$(x) = 5x^2 - 4x + 12$$ leads to the coefficient of $$10$$’’ in $$p'(x) = 10x -4\text{.}$$

#### 1.4.3.8.

1. $$g$$ is linear.
2. On $$-3.5 \lt x \lt -2\text{,}$$ $$-2 \lt x \lt 0$$ and $$2 \lt x \lt 3.5\text{.}$$
3. At $$x = -2, 0, 2\text{;}$$ $$g$$ must have sharp corners at these points.

### 1.5Interpreting, estimating, and using the derivative1.5.4Exercises

#### 1.5.4.5.

1. $$F'(10) \approx -3.33592\text{.}$$
2. The coffee’s temperature is decreasing at about $$3.33592$$ degrees per minute.
3. $$F'(20)\text{.}$$
4. We expect $$F'$$ to get closer and closer to $$0$$ as time goes on.

#### 1.5.4.6.

1. If a patient takes a dose of $$50$$ ml of a drug, the patient will experience a body temperature change of $$0.75$$ degrees F.
2. degrees Fahrenheit per milliliter.’’
3. For a patient taking a $$50$$ ml dose, adding one more ml to the dose leads us to expect a temperature change that is about $$0.02$$ degrees less than the temperature change induced by a $$50$$ ml dose.

#### 1.5.4.7.

1. $$t=0\text{.}$$
2. $$v'(1) = -32\text{.}$$
3. feet per second per second’’; $$v'(1) = -32$$ tells us that the ball’s velocity is decreasing at a rate of 32 feet per second per second.
4. The acceleration of the ball.

#### 1.5.4.8.

1. $$AV_{[40000,55000]} \approx -0.153$$ dollars per mile.
2. $$h'(55000) \approx -0.147$$ dollars per mile. During $$55 0001$$st mile, we expect the car’s value to drop by $$0.147$$ dollars.
3. $$h'(30000) \lt h'(80000)\text{.}$$
4. The graph of $$h$$ might have the general shape of the graph of $$y = e^{-x}$$ for positive values of $$x\text{:}$$ always positive, always decreasing, and bending upwards while tending to $$0$$ as $$x$$ increases.

### 1.6The second derivative1.6.5Exercises

#### 1.6.5.6.

1. $$f$$ is increasing and concave down near $$x=2\text{.}$$
2. Greater.
3. Less.

#### 1.6.5.7.

1. $$g'(2) \approx 1.4\text{.}$$
2. At most one.
3. $$9\text{.}$$
4. $$g''(2) \approx 5.5 \text{.}$$

#### 1.6.5.8.

1. $$h'(4.5) \approx 14.3\text{;}$$ $$h'(5) \approx 21.2\text{;}$$ $$h'(5.5) \approx = 23.9\text{;}$$ rising most rapidly at $$t = 5.5\text{.}$$
2. $$h'(5) \approx 9.6 \text{.}$$
3. Acceleration of the bungee jumper in feet per second per second.
4. $$0 \lt t \lt 2\text{,}$$ $$6 \lt t \lt 10\text{.}$$

### 1.7Limits, Continuity, and Differentiability1.7.5Exercises

#### 1.7.5.5.

1. $$a = 0\text{.}$$
2. $$a = 0, 3\text{.}$$
3. $$a = -2, 0, 1, 2, 3\text{.}$$

#### 1.7.5.6.

1. $$f(x) = |x-2|\text{.}$$
2. Impossible.
3. Let $$f$$ be the function defined to be $$f(x) = 1$$ for every value of $$x \ne -2\text{,}$$ and such that $$f(-2) = 4\text{.}$$

#### 1.7.5.7.

1. $$h$$ must be piecewise linear with slope of $$1$$ or $$-1\text{,}$$ depending on the interval.
2. $$h'(x)$$ is not defined for $$x = -2, 0, 2\text{.}$$
3. It is possible that $$h$$ is not continuous at $$x = -2, 0, 2\text{.}$$
4. Two of the many possible graphs for $$h$$ are shown in the following figure.

#### 1.7.5.8.

1. At $$x = 0\text{.}$$
\begin{align*} g'(0) & = \lim_{h \to 0} \frac{g(0+h) - g(0)}{h}\\ & = \lim_{h \to 0} \frac{\sqrt{|h|} - \sqrt{|0|}}{h}\\ & = \lim_{h \to 0} \frac{\sqrt{|h|}}{h} \end{align*}
2.  $$h$$ $$0.1$$ $$0.01$$ $$0.001$$ $$0.0001$$ $$-0.1$$ $$-0.01$$ $$-0.001$$ $$-0.0001$$ $$\sqrt{|h|}/h$$ $$3.162$$ $$10$$ $$31.62$$ $$100$$ $$-3.162$$ $$-10$$ $$-31.62$$ $$-100$$
$$g'(0)$$ does not exist.

### 1.8The Tangent Line Approximation1.8.4Exercises

#### 1.8.4.5.

1. $$p(3) = -1$$ and $$p'(3) = -2\text{.}$$
2. $$p(2.79) \approx -0.58\text{.}$$
3. Too large.

#### 1.8.4.6.

1. $$F'(60)\approx 1.56$$ degrees per minute.
2. $$L(t) \approx 1.56(t-60)+324.5\text{.}$$
3. $$F(63)\approx L(63)\approx = 329.18$$ degrees F.
4. Overestimate.

#### 1.8.4.7.

1. $$s(9.34) \approx L(9.34) = 3.592\text{.}$$
2. underestimate.
3. The object is slowing down as it moves toward toward its starting position at $$t=4\text{.}$$

#### 1.8.4.8.

1. $$x=1\text{.}$$
2. On $$-0.37 \lt x \lt 1.37\text{;}$$ $$f$$ is concave up.
3. $$f(1.88) \approx -3.0022\text{,}$$ and this estimate is larger than the true value of $$f(1.88)\text{.}$$

### 2Computing Derivatives2.1Elementary derivative rules2.1.5Exercises

#### 2.1.5.10.

1. $$h(2) = 27\text{;}$$ $$h'(2) = -19/2\text{.}$$
2. $$L(x) = 27 - \frac{19}{2}(x-2)\text{.}$$
3. $$p$$ is increasing at $$x=2\text{.}$$
4. $$p(2.03) \approx -11.44\text{.}$$

#### 2.1.5.11.

1. $$p$$ is not differentiable at $$x=-1$$ and $$x=1\text{;}$$ $$q$$ is not differentiable at $$x=-1$$ and $$x=1\text{.}$$
2. $$r$$ is not differentiable at $$x=-1$$ and $$x=1\text{.}$$
3. $$r'(-2) = 4\text{;}$$ $$r'(0) = \frac{1}{2}\text{.}$$
4. $$y = 4\text{.}$$

#### 2.1.5.12.

1. $$w'(t) = 3t^t(\ln(t) + 1) + 2\frac{1}{\sqrt{1-t^2}}\text{.}$$
2. $$L(t) = (\frac{3}{\sqrt{2}} - \frac{2\pi}{3}) + (\frac{3}{\sqrt{2}}(\ln(\frac{1}{2}) + 1) + \frac{4}{\sqrt{3}})(t-\frac{1}{2})\text{.}$$
3. $$v$$ is decreasing at $$t = \frac{1}{2}\text{.}$$

#### 2.1.5.13.

1. \begin{align*} f'(x) &= \lim_{h \to 0} \frac{a^{x+h}-a^x}{h}\\ &= \lim_{h \to 0} \frac{a^{x} \cdot a^{h} - a^x}{h}\\ &= \lim_{h \to 0} \frac{a^{x}(a^{h}-1)}{h}\text{,} \end{align*}
2. Since $$a^x$$ does not depend at all on $$h\text{,}$$ we may treat $$a^x$$ as constant in the noted limit and thus write the value $$a^x$$ in front of the limit being taken.
3. When $$a = 2\text{,}$$ $$L \approx 0.6931\text{;}$$ when $$a = 3\text{,}$$ $$L \approx 1.0986\text{.}$$
4. $$a \approx 2.71828$$ (for which $$L \approx 1.000$$)
5. $$\frac{d}{dx}[2^x] = 2^x \cdot \ln(2)$$ and $$\frac{d}{dx}[3^x] = 3^x \cdot \ln(3)$$
6. \begin{equation*} \frac{d}{dx}[e^x] = e^x\text{.} \end{equation*}

### 2.2The sine and cosine functions2.2.3Exercises

#### 2.2.3.1.

1. $$V'(2) = 24 \cdot 1.07^2 \cdot \ln(1.07) + 6 \cos(2) \approx -0.63778$$ thousands of dollars per year.
2. $$V''(2) = 24 \cdot 1.07^2 \cdot \ln(1.07)^2 - 6 \sin(2) \approx -5.33$$ thousands of dollars per year per year. At this moment, $$V'$$ is decreasing and we expect the derivative’s value to decrease by about $$5.33$$ thousand dollars per year over the course of the next year.
3. See the figure below. Adding the term $$6\sin(t)$$ to $$A$$ to create the function $$V$$ adds volatility to the value of the portfolio.

#### 2.2.3.2.

1. $$f'\left(\frac{\pi}{4}\right) = -5\left(\frac{\sqrt{2}}{2}\right)\text{.}$$
2. $$L(x) = 3+2(x-\pi)\text{.}$$
3. Decreasing.
4. The tangent line to $$f$$ lies above the curve at this point.

#### 2.2.3.3.

1. Hint: in the numerator of the difference quotient, combine the first and last terms and remove a factor of $$\sin(x)\text{.}$$
2. Hint: divide each part of the numerator by $$h$$ and consider the sum of two separate limits.
3. $$\lim_{h \to 0} \left( \frac{\cos(h)-1}{h} \right) = 0$$ and $$\lim_{h \to 0} \left( \frac{\sin(h)}{h} \right) = 1 \text{.}$$
4. $$f'(x) = \sin(x) \cdot 0 + \cos(x) \cdot 1\text{.}$$
5. Hint: $$\cos(\alpha + \beta)$$ is $$\cos(\alpha + \beta) = \cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta)\text{.}$$

### 2.3The product and quotient rules2.3.5Exercises

#### 2.3.5.10.

1. $$h(2) = -15\text{;}$$ $$h'(2) = 23/2\text{.}$$
2. $$L(x) = -15 + 23/2(x-2)\text{.}$$
3. Increasing.
4. $$r(2.06) \approx -0.5796\text{.}$$

#### 2.3.5.11.

1. $$w'(t) = t^t\left(\ln t+1\right)\cdot\left(\arccos t\right)+t^t\cdot\frac{-1}{\sqrt{1-t^2}} \text{.}$$
2. $$L(t) \approx 0.740-0.589(t-0.5)\text{.}$$
3. Increasing.

#### 2.3.5.12.

1. $$r'(-2) = 5$$ and $$r'(0) = 1\text{.}$$
2. At $$x = -1$$ and $$x = 1\text{.}$$
3. $$L(x) = 2\text{.}$$
4. $$z'(0) = -\frac{1}{4}$$ and $$z'(2) = -1\text{.}$$
5. At $$x = -1\text{,}$$ $$x = 1\text{,}$$ $$x = -1.5\text{,}$$ and $$x = 1\text{.}$$

#### 2.3.5.13.

1. $$C(t) = A(t)Y(t)$$ bushels in year $$t\text{.}$$
2. $$1 190 000$$ bushels of corn.
3. $$C'(t) = A(t)Y'(t) + A'(t)Y(t)\text{.}$$
4. $$C'(0) = 158 000$$ bushels per year.
5. $$C(1) \approx 1 348 000$$bushels.

#### 2.3.5.14.

1. $$g(80) = 20$$ kilometers per liter, and $$g'(80) = -0.16\text{.}$$ kilometers per liter per kilometer per hour.
2. $$h(80) = 4$$ liters per hour and $$h'(80) = 0.082$$ liters per hour per kilometer per hour.
3. Think carefully about units and how each of the three pairs of values expresses fundamentally the same facts.

### 2.4Derivatives of other trigonometric functions2.4.3Exercises

#### 2.4.3.6.

1. $$h'(2) = \frac{-2\sin(2) - 2\cos(2) \ln(1.2)}{1.2^2} \approx -1.1575$$ feet per second.
2. $$h''(2) = \frac{\cos(2)(-2 + 2ln^2(1.2)) + 4\ln(1.2)\sin(2))}{1.2^2} \approx 1.0193$$ feet per second per second.
3. The object is falling and slowing down.

#### 2.4.3.7.

1. $$f'(x) = \sin(x) \cdot (-\csc^2(x)) + \cot(x) \cdot \cos(x)\text{.}$$
2. False.
3. $$f'(x) = \frac{-\sin^2(x)}{\sin(x)} = -\sin(x)$$ for $$x \ne \frac{\pi}{2} + k\pi$$ for some integer value of $$k\text{.}$$

#### 2.4.3.8.

1. $$\displaystyle p'(z) = \frac{\left(z^2\sec(z) +1 \right)\left(z\sec^2(z)+\tan(z)\right) - z\tan(z) \left(z^2\sec(z)\tan(z)+2z\sec(z)\right)}{\left(z^2\sec(z) + 1\right)^2} +3e^z$$
2. $$y - 4 = 3(x-0)\text{.}$$
3. Increasing.

### 2.5The chain rule2.5.5Exercises

#### 2.5.5.8.

1. $$h'\left( \frac{\pi}{4} \right) = \frac{3}{2\sqrt{2}}\text{.}$$
2. $$r'(0.25) = \cos(0.25^3) \cdot 3(0.25)^2 \approx 0.1875 \gt h'(0.25) = 3\sin^2(0.25) \cdot \cos(0.25) \approx 0.1779\text{;}$$ $$r$$ is changing more rapidly.
3. $$h'(x)$$ is periodic; $$r'(x)$$ is not.

#### 2.5.5.9.

1. $$p'(x) = e^{u(x)} \cdot u'(x)\text{.}$$
2. $$q'(x) = u'(e^x) \cdot e^x\text{.}$$
3. $$r'(x) = -\csc^2(u(x)) \cdot u'(x)\text{.}$$
4. $$s'(x) = u'(\cot(x)) \cdot (-\csc^2(x))\text{.}$$
5. $$a'(x) = u'(x^4) \cdot 4x^3\text{.}$$
6. $$b'(x) = 4(u(x))^3 \cdot u'(x)\text{.}$$

#### 2.5.5.10.

1. $$C'(0) = 0$$ and $$C'(3) = -\frac{1}{2}\text{.}$$
2. Consider $$C'(1)\text{.}$$ By the chain rule, we’d expect that $$C'(1) = p'(q(1)) \cdot q'(1)\text{,}$$ but we know that $$q'(1)$$ does not exist since $$q$$ has a corner point at $$x = 1\text{.}$$ This means that $$C'(1)$$ does not exist either.
3. Since $$Y(x) = q(q(x))\text{,}$$ the chain rule implies that $$Y'(x) = q'(q(x)) \cdot q'(x)\text{,}$$ and thus $$Y'(-2) = q'(q(-2)) \cdot q'(-2) = q'(-1) \cdot q'(-2)\text{.}$$ But $$q'(-1)$$ does not exist, so $$Y'(-2)$$ also fails to exist. Using $$Z(x) = q(p(x))$$ and the chain rule, we have $$Z'(x) = q'(p(x)) \cdot p'(x)\text{.}$$ Therefore $$Z'(0) = q'(p(0)) \cdot p'(0) = q'(-0.5) \cdot p'(0) = 0 \cdot 0.5 = 0\text{.}$$

#### 2.5.5.11.

1. $$\left. \frac{dV}{dh} \right|_{h=1} = 7 \pi$$ cubic feet per foot.
2. $$h'(2) = \pi \cos(2\pi) = \pi$$ feet per hour.
3. $$\left. \frac{dV}{dt} \right|_{t=2} = 7 \pi^2$$ cubic feet per hour.
4. In (a) we are determining the instantaneous rate at which the volume changes as we increase the height of the water in the tank, while in (c) we are finding the instantaneous rate at which volume changes as we increase time.

### 2.6Derivatives of Inverse Functions2.6.6Exercises

#### 2.6.6.9.

1. $$f'(x) = \frac{1}{2\arctan(x) + 3\arcsin(x) + 5} \cdot \left(\frac{2}{1+x^2} + \frac{3}{\sqrt{1-x^2}}\right)\text{.}$$
2. $$r'(z) = \frac{1}{1+\left(\ln(\arcsin(z))\right)^2} \cdot \left( \frac{1}{\arcsin(z)} \right) \cdot \frac{1}{\sqrt{1-z^2}}\text{.}$$
3. $$q'(t) = \arctan^2(3t) \cdot \left[4\arcsin^3(7t) \left( \frac{7}{\sqrt{1-(7t)^2}} \right)\right] + \arcsin^4(7t) \cdot \left[2\arctan(3t) \left(\frac{3}{1+(3t)^2}\right) \right]\text{.}$$
4. $$\displaystyle g'(v) = \frac{1}{\frac{\arctan(v)}{\arcsin(v) + v^2}} \cdot \frac{(\arcsin(v) + v^2) \cdot \frac{1}{1+v^2} - \arctan(v) \cdot \left(\frac{1}{\sqrt{1-v^2}} + 2v \right)}{(\arcsin(v) + v^2)^2}$$

#### 2.6.6.10.

1. $$f'(1) \approx 2\text{.}$$
2. $$(f^{-1})'(-1) \approx 1/2\text{.}$$

#### 2.6.6.11.

1. $$f$$ passes the horizontal line test.
2. $$f^{-1}(x) = g(x) = \sqrt[3]{4x-16}\text{.}$$
3. $$f'(x) = \frac{3}{4}x^2\text{;}$$ $$f'(2) = 3\text{.}$$ $$g'(x) = \frac{1}{3}(4x-16)^{-2/3} \cdot 4\text{;}$$ $$g'(6) = \frac{1}{3}\text{.}$$ These two derivative values are reciprocals.

#### 2.6.6.12.

1. $$h$$ passes the horizontal line test.
2. The equation $$y = x + \sin(x)$$ can’t be solved for $$x$$ in terms of $$y\text{.}$$
3. $$(h^{-1})'(\frac{\pi}{2} + 1) = 1\text{.}$$

### 2.7Derivatives of Functions Given Implicitly2.7.3Exercises

#### 2.7.3.6.

Horizontal tangent lines: $$(0,-1)\text{,}$$ $$(0,-0.618)\text{,}$$ $$(0,1.618)\text{,}$$ $$(1,-1)\text{,}$$ $$(1,-0.618)\text{,}$$ $$(1,1.618)\text{,}$$ $$(0.5,-1.0493)\text{,}$$ $$(0.5,0.2104)\text{,}$$ $$(0.5, 1.6139)\text{.}$$ Vertical tangent lines: $$(-0.1756,-0.379)\text{,}$$ $$(0.2912,-0.379)\text{,}$$ $$(0.7088,-0.379)\text{,}$$ $$(1.1756,-0.379)\text{,}$$ $$(-0.8437, 1.235)\text{,}$$ and $$(1.8437, 1.235)\text{.}$$

#### 2.7.3.7.

$$y = \frac{\pi}{2} - \left(x-\frac{\pi}{2}\right)\text{.}$$

#### 2.7.3.8.

1. $$x = \frac{\ln(y)}{\ln(a)}\text{.}$$
2. $$1 = \frac{1}{\ln(a)} \cdot \frac{1}{y}\frac{dy}{dx}\text{.}$$
3. $$\frac{d}{dx}[a^x] = a^x \ln(a)\text{.}$$

### 2.8Using Derivatives to Evaluate Limits2.8.4Exercises

#### 2.8.4.5.

$$\lim_{x \to 3} h(x) = -2\text{.}$$

#### 2.8.4.6.

Horizontal asymptote: $$y = \frac{3}{5}\text{;}$$ vertical asymptote: $$x = c\text{;}$$ hole: $$(a, \frac{3(a-b)}{5(a-c)})\text{.}$$ $$R$$ is not continuous at $$x = a$$ and $$x = c\text{.}$$

#### 2.8.4.7.

1. $$\ln(x^{2x}) = 2x \cdot \ln(x)\text{.}$$
2. $$x = \frac{1}{\frac{1}{x}}\text{.}$$
3. $$\lim_{x \to 0^+} h(x) = 0\text{.}$$
4. $$\lim_{x \to 0^+} g(x) = \lim_{x \to 0^+} x^{2x} = 1\text{.}$$

#### 2.8.4.8.

1. Show that $$\lim_{x \to \infty}\frac{\ln(x)}{\sqrt{x}} = 0\text{.}$$
2. Show that $$\lim_{x \to \infty}\frac{\ln(x)}{\sqrt[n]{x}} = 0\text{.}$$
3. Consider $$\lim_{x \to \infty} \frac{p(x)}{e^x}$$ By repeated application of LHR, the numerator will eventually be simply a constant (after $$n$$ applications of LHR), and thus with $$e^x$$ still in the denominator, the overall limit will be $$0\text{.}$$
4. Show that $$\lim_{x \to \infty} \frac{\ln(x)}{x^n} = 0$$
5. For example, $$f(x) = 3x^2 + 1$$ and $$g(x) = -0.5x^2 + 5x - 2\text{.}$$

### 3Using Derivatives3.1Using derivatives to identify extreme values3.1.4Exercises

#### 3.1.4.4.

1. $$f'$$ is positive for $$-1 \lt x lt 1$$ and for $$x \gt 1\text{;}$$ $$f'$$ is negative for all $$x \lt -1\text{.}$$ $$f$$ has a local minimum at $$x = -1\text{.}$$
2. A possible graph of $$y = f''(x)$$ is shown at right in the figure.
3. $$f''(x)$$ is negative for $$-0.35 \lt x \lt 1\text{;}$$ $$f''(x)$$ is positive everywhere else; $$f$$ has points of inflection at $$x \approx -0.35$$ and $$x = 1\text{.}$$
4. A possible graph of $$y = f(x)$$ is shown at left in the figure.

#### 3.1.4.5.

1. Neither.
2. $$g''(2) = 0\text{;}$$ $$g''$$ is negative for $$1 \lt x \lt 2$$ and positive for $$2 \lt x \lt 3\text{.}$$
3. $$g$$ has a point of inflection at $$x = 2\text{.}$$

#### 3.1.4.6.

1. $$h$$ can have no, one, or two real zeros.
2. One root is negative and the other positive.
3. $$h$$ will look like a line with slope $$3\text{.}$$
4. $$h$$ is concave up everywhere; $$h$$ is almost linear for large values of $$|x|\text{.}$$

#### 3.1.4.7.

1. $$p''(x)$$ is negative for $$-1 \lt x \lt 2$$ and positive for all other values of $$x\text{;}$$ $$p$$ has points of inflection at $$x = -1$$ and $$x = 2\text{.}$$
2. Local maximum.
3. Neither.

### 3.2Using derivatives to describe families of functions3.2.3Exercises

#### 3.2.3.3.

1. $$x = 0$$ and $$x = \frac{2a}{3}\text{.}$$
2. $$x = \frac{a}{3}\text{;}$$ $$p''(x)$$ changes sign from negative to positive at $$x = \frac{a}{3}\text{.}$$
3. As we increase the value of $$a\text{,}$$ both the location of the critical number and the inflection point move to the right along with $$a\text{.}$$

#### 3.2.3.4.

1. $$x=c$$ is a vertical asymptote because $$\lim_{x \to c^+} \frac{e^{-x}}{x-c} = \infty$$ and $$\lim_{x \to c^-} \frac{e^{-x}}{x-c} = -\infty\text{.}$$
2. $$\lim_{x \to \infty} \frac{e^{-x}}{x-c} = 0\text{;}$$ $$\lim_{x \to -\infty} \frac{e^{-x}}{x-c} = -\infty\text{.}$$
3. The only critical number for $$q$$ is $$x=c-1\text{.}$$
4. When $$x \lt c-1\text{,}$$ $$q'(x) \gt 0\text{;}$$ when $$x \gt c-1\text{,}$$ $$q'(x) \lt 0\text{;}$$ $$q$$ has a local maximum at $$x = c-1\text{.}$$

#### 3.2.3.5.

1. $$x = m\text{.}$$
2. $$E$$ is increasing for $$x \lt m$$ and decreasing for $$x \gt m\text{,}$$ with a local maximum at $$x = m\text{.}$$
3. $$x = m \pm s\text{.}$$
4. $$\lim_{x \to \infty} E(x) = \lim_{x \to -\infty} E(x) = 0\text{.}$$

### 3.3Global Optimization3.3.4Exercises

#### 3.3.4.1.

1. Not enough information is given.
2. Global minimum at $$x = b\text{.}$$
3. Global minimum at $$x = a\text{;}$$ global maximum at $$x = b\text{.}$$
4. Not enough information is provided.

#### 3.3.4.2.

1. Absolute maximum $$p(0) = p(a) = 0\text{;}$$ absolute minimum $$p\left( \frac{a}{\sqrt{3}} \right) = -\frac{2a^3}{3\sqrt{3}}\text{.}$$
2. Absolute max $$r\left( \frac{1}{b} \right) \approx 0.368 \frac{a}{b}\text{;}$$ absolute min $$r\left( \frac{2}{b} \right) \approx 0.270 \frac{a}{b}\text{.}$$
3. Absolute minimum $$g(b) = a(1-e^{-b^2})\text{;}$$ absolute maximum $$g(3b) = a(1-e^{-3b^2})\text{.}$$
4. Absolute max $$s\left( \frac{\pi}{2k} \right) = 1\text{;}$$ absolute min $$s\left( \frac{5\pi}{6k} \right) = \frac{1}{2}\text{.}$$

#### 3.3.4.3.

1. Global maximum at $$x=a\text{;}$$ global minimum at $$x=b\text{.}$$
2. Global maximum at $$x=c\text{;}$$ global minimum at either $$x=a$$ or $$x=b\text{.}$$
3. Global minimum at $$x=a$$ and $$x=b\text{;}$$ global maximum somewhere in $$(a,b)\text{.}$$
4. Global minimum at $$x=c\text{;}$$ global maximum value at $$x = a\text{.}$$

#### 3.3.4.4.

1. Absolute max $$s(\frac{5\pi}{12}) = 8\text{;}$$ absolute min $$s(\frac{11\pi}{12}) = 2\text{.}$$
2. Absolute max $$s(\frac{5\pi}{12}) = 8\text{;}$$ absolute min $$s(0) = 5 - \frac{3\sqrt{3}}{2} \approx 2.402\text{.}$$
3. Absolute max $$s(\frac{5\pi}{12}) = 8\text{;}$$ absolute min $$s(\frac{11\pi}{12}) = 2\text{.}$$ (There are other points at which the function achieves these values on the given interval.)
4. Absolute max $$s(\frac{5\pi}{12}) = 8\text{;}$$ absolute min $$s(\frac{5\pi}{6}) \approx 2.402\text{.}$$

### 3.4Applied Optimization3.4.3Exercises

#### 3.4.3.6.

The absolute maximum volume is $$V\left( \sqrt{\frac{15}{18}} \right) = \frac{15}{6}\left( \sqrt{\frac{15}{18}} \right) - \left( \sqrt{\frac{15}{18}} \right)^3 \approx 1.52145$$ cubic feet.

#### 3.4.3.7.

The maximum possible area that each of the four pens can enclose is 351562.5 square feet.

#### 3.4.3.8.

$$172.047$$ feet of cable.

### 8.3Series of Real Numbers8.3.7Exercises

#### 8.3.7.5.

1. $$\sum \frac{10^k}{k!}$$ converges.
2. $$\sum \frac{10^k}{k!}$$ converges.
3. The sequence $$\left\{\frac{b^n}{n!}\right\}$$ has to converge to 0.

#### 8.3.7.6.

1. $$\sqrt[n]{a_n} \approx r$$ for large $$n\text{.}$$
2. $$\frac{a_{n+1}}{a_n} \approx r\text{.}$$
3. $$0 \lt r \lt 1\text{.}$$

#### 8.3.7.7.

1. $$a_n = 1 + \frac{1}{2^n} \to 1 \ne 0$$ and $$b_n = -1 \to -1 \ne 0\text{.}$$
2. The series is geometric with $$r = \frac{1}{2}\text{.}$$
3. Since the two individual series diverge, neither sum is a finite number, so it doesn’t make any sense to add them.
1. Note that $$A_n + B_n = (a_1 + b_1) + (a_2 + b_2) + \cdots + (a_n + b_n) \text{.}$$
2. Note that $$\lim_{n \to \infty} \sum_{k=1}^n (a_k+b_k) = \lim_{n \to \infty} \left( \sum_{k=1}^{n} a_k + \sum_{k=1}^{n} b_k \right) \text{.}$$
1. $$\sum_{k=0}^{\infty} \frac{2^k+3^k}{5^k} = \frac{25}{6} \text{.}$$

#### 8.3.7.8.

1. $$S_1 = 1$$ and $$T_1 = \frac{1}{2}\text{.}$$
2. $$S_2 \gt T_2\text{.}$$
3. $$S_3 \gt T_3\text{.}$$
4. $$S_n \gt T_n\text{.}$$
5. $$\sum \frac{1}{k^2} \gt \sum \frac{1}{k^2+k} \text{;}$$ $$\sum \frac{1}{k^2+k}$$ converges.
1. If $$\sum b_k$$ diverges, then $$\sum b_k$$ is infinite, and anything larger must also be infinite; if $$\sum a_k$$ is convergent then anything smaller and positive must also be finite.
1. Note that $$0 \lt \frac{1}{k} \lt \frac{1}{k-1}\text{.}$$
2. Note that $$\frac{1}{k^3} \gt \frac{1}{k^3+1}\text{.}$$

### 8.4Alternating Series8.4.6Exercises

#### 8.4.6.5.

1. $$\sum_{k=0}^{\infty} \frac{1}{2k+1}$$ diverges by comparison to the Harmonic series.
2. $$|S_{100} - \sum_{k=0}^{\infty} (-1)^{k} \frac{1}{2k+1}| \lt \approx 0.0049 \text{.}$$
3. $$\displaystyle n \gt 4{,}999{,}999{,}999.5$$

#### 8.4.6.6.

1. $$\frac{S_n+S_{n+1}}{2} = \frac{S_n+S_{n} + (-1)^{n+2}a_{n+1}}{2} \text{.}$$
2. $$S_{20} = 0.668771403 \ldots \text{;}$$ $$\frac{S_{20} + S_{21}}{2} = \frac{161227687}{232792560} = 0.692580926 \ldots \text{,}$$ accurate to within about $$0.0006\text{.}$$

#### 8.4.6.7.

1. $$\left\{\frac{1}{n}\right\}$$ and $$\left\{-\frac{1}{n^2}\right\}$$ converge to 0.
2. Notice that $$\frac{1}{k} - \frac{1}{k^2} = \frac{k-1}{k^2}$$ and compare to the Harmonic series.
3. It is possible for a series to alternate, have the terms go to zero, have the terms not decrease to zero, and the series diverge.

### 8.5Taylor Polynomials and Taylor Series8.5.6Exercises

#### 8.5.6.6.

1. $$P_3(x) = -1 + 3x - \frac{4}{2!}x^2 + \frac{6}{3!}x^3 \text{,}$$ which is the same polynomial as $$f(x)\text{.}$$
2. For $$n \ge 3\text{,}$$ $$P_n(x) = f(x)\text{.}$$
3. For $$n \ge m\text{,}$$ $$P_n(x) = f(x)\text{.}$$

#### 8.5.6.7.

1. \begin{align*} P_1(x) &= 1 + 0\left(x - \frac{\pi}{2} \right) = 1\\ P_2(x) &= 1 + 0\left(x - \frac{\pi}{2} \right) - \frac{1}{2!}\left(x - \frac{\pi}{2} \right)^2 = 1-\frac{1}{2!}\left(x - \frac{\pi}{2} \right)^2\\ P_3(x) &= 1 + 0\left(x - \frac{\pi}{2} \right) - \frac{1}{2!}\left(x - \frac{\pi}{2} \right)^2 + \frac{0}{3!}\left(x - \frac{\pi}{2} \right)^3\\ &= 1-\frac{1}{2!}\left(x - \frac{\pi}{2} \right)^2 = P_2(x)\\ P_4(x) &= 1 + 0\left(x - \frac{\pi}{2} \right) - \frac{1}{2!}\left(x - \frac{\pi}{2} \right)^2 + \frac{0}{3!}\left(x - \frac{\pi}{2} \right)^3 + \frac{1}{4!}\left(x - \frac{\pi}{2} \right)^4\\ &= 1 - \frac{1}{2!}\left(x - \frac{\pi}{2} \right)^2 + \frac{1}{4!}\left(x - \frac{\pi}{2} \right)^4\\ P(x) &= 1 - \frac{1}{2!}\left(x - \frac{\pi}{2} \right)^2 + \frac{1}{4!}\left(x - \frac{\pi}{2} \right)^4 - \frac{1}{6!}\left(x - \frac{\pi}{2} \right)^6 + \cdots \end{align*}
2. \begin{equation*} P_4(x) = 0 + 1(x-1) - \frac{1}{2!}(x-1)^2 + \frac{2}{3!}(x-1)^3 - \frac{6}{4!}(x-1)^4\text{.} \end{equation*}
\begin{equation*} P(x) = 1(x-1) - \frac{1}{2}(x-1)^2 + \frac{1}{3}(x-1)^3 - \frac{1}{4}(x-1)^4 + \frac{1}{5}(x-1)^5 - \cdots \end{equation*}
3. $$\displaystyle P_{101}(1) \approx 0.698073$$

#### 8.5.6.8.

1. $$P_4(x) = x^2 \text{.}$$
1. $$g(x) = P(x^2) = x^2 - \frac{1}{3!}x^6 + \frac{1}{5!}x^10 - \cdots \text{.}$$
2. All real numbers.

### 8.6Power Series8.6.4Exercises

#### 8.6.4.3.

1. $$\sin(x^2) = \sum_{k=0}^{\infty} (-1)^k\frac{x^{2(2k+1)}}{(2k+1)!} \text{,}$$ with interval of convergence $$(-\infty, \infty)\text{.}$$
2. $$\int \sin(x^2) \, dx = \sum_{k=0}^{\infty} (-1)^k\frac{x^{4k+3}}{(2k+1)!(4k+3)} + C \text{.}$$
3. $$\int_0^1 \sin(x^2) \, dx = \sum_{k=0}^{\infty} (-1)^k\frac{1}{(2k+1)!(4k+3)} \text{.}$$ Use $$n = 1$$ to generate the desired estimate.

#### 8.6.4.4.

for each $$k \geq 0\text{.}$$ But these are just the coefficients of the Taylor series expansion of $$f\text{,}$$ which leads us to the following observation.