## 17Calculus - Calculus 1 - Practice Exam 2 (Final) (Semester A)

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This is the final exam for first semester single variable calculus.

### Practice Exam Tips

Each exam page contains a full exam with detailed solutions. Most of these are actual exams from previous semesters used in college courses. You may use these as practice problems or as practice exams. Here are some suggestions on how to use these to help you prepare for your exams.

- Set aside a chunk of full, uninterrupted time, usually an hour to two, to work each exam.
- Go to a quiet place where you will not be interrupted that duplicates your exam situation as closely as possible.
- Use the same materials that you are allowed in your exam (unless the instructions with these exams are more strict).
- Use your calculator as little as possible except for graphing and checking your calculations.
- Work the entire exam before checking any solutions.
- After checking your work, rework any problems you missed and go to the 17calculus page discussing the material to perfect your skills.
- Work as many practice exams as you have time for. This will give you practice in important techniques, experience in different types of exam problems that you may see on your own exam and help you understand the material better by showing you what you need to study.

IMPORTANT -
Exams can cover only so much material. Instructors will sometimes change exams from one semester to the next to adapt an exam to each class depending on how the class performs during the semester while they are learning the material. So just because you do well (or not) on these practice exams, does not necessarily mean you will do the same on your exam. Your instructor may ask completely different questions from these. That is why working lots of practice problems will prepare you better than working just one or two practice exams.
Calculus is not something that can be learned by reading. You have to work problems on your own and struggle through the material to really know calculus, do well on your exam and be able to use it in the future.

Exam Details

Time

80 minutes

Questions

14

Total Points

100

Tools

Calculator

see instructions

Formula Sheet(s)

none

Other Tools

none

Instructions:
- This exam is in two main parts, labeled parts A and B, with different instructions for each part.
- Show all your work.
- For each problem, correct answers are worth 1 point. The remaining points are earned by showing calculations and giving reasoning that justify your conclusions.
- Correct notation counts (i.e. points will be taken off for incorrect notation).
- Give exact, simplified answers.

Part A - Questions 1-8

Instructions for Part A - - You have 40 minutes to complete this part of the exam. No calculators are allowed. Each question in this section is worth 5 points.

Section 1

Evaluate the following integrals.

$$\displaystyle{ \int_1^3{ \frac{5x+5}{x^2+2x+2} ~dx } }$$

Problem Statement

$$\displaystyle{ \int_1^3{ \frac{5x+5}{x^2+2x+2} ~dx } }$$

$$\displaystyle{ \frac{5}{2}\ln\left[ \frac{17}{5} \right] }$$

Problem Statement

$$\displaystyle{ \int_1^3{ \frac{5x+5}{x^2+2x+2} ~dx } }$$

Solution

Let's try integration by substitution.

 Let $$u = x^2 + 2x + 2$$ $$du = (2x+2)dx$$ $$\displaystyle{ \frac{du}{2(x+1)} = dx }$$ We will solve the indefinite integral first. $$\displaystyle{ \int{ \frac{5x+5}{x^2+2x+2} ~dx } }$$ Applying the substitution we chose into the integral gives us $$\displaystyle{ \int{ \frac{5(x+1)}{u} \frac{du}{2(x+1)} } }$$ Cancel the $$(x+1)$$ terms in the numerator and denominator. $$\displaystyle{ \frac{5}{2}\int{ \frac{du}{u} } }$$ $$\displaystyle{ \frac{5}{2} \ln|u|}$$ We are solving an indefinite integral, so we should have a constant of integration. However, looking ahead, our problem is a definite integral which will cause the constant to drop out soon. So we do not need it for this problem. Now we have $$\displaystyle{ \frac{5}{2} \ln|x^2+2x+2| }$$ Now we apply the limits of integration. $$\displaystyle{ \left[ \frac{5}{2} \ln|x^2+2x+2| \right]_1^3 }$$ $$\displaystyle{ \frac{5}{2}\left[ \ln|9+6+2| - \ln|1+2+2 \right] }$$ $$\displaystyle{ \frac{5}{2} \left[ \ln 17 - \ln 5 \right] }$$ $$\displaystyle{ \frac{5}{2}\ln\left[ \frac{17}{5} \right] }$$

$$\displaystyle{ \frac{5}{2}\ln\left[ \frac{17}{5} \right] }$$

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$$\displaystyle{ \int_{-1}^1{ x \cos(x^3) ~dx } }$$

Problem Statement

$$\displaystyle{ \int_{-1}^1{ x \cos(x^3) ~dx } }$$

$$\displaystyle{ \int_{-1}^1{ x \cos(x^3) ~dx } = 0 }$$

Problem Statement

$$\displaystyle{ \int_{-1}^1{ x \cos(x^3) ~dx } }$$

Solution

This problem has a unique solution. So we will go through several techniques to see what won't work and help understand how to solve this problem. First, let's try integration by substitution.

 Let $$u = x^3$$. $$du = 3x^2 ~dx$$ $$\displaystyle{ \int{ x \cos(x^3) ~dx } }$$ $$\displaystyle{ \int{ x \cos(u) \frac{du}{3x^2} } }$$ $$\displaystyle{ \frac{1}{3} \int{ \frac{\cos(u)}{u^{1/3}} ~du } }$$ This integral does not seem to have an easy solution. So let's start over trying integration by parts. $$u = \cos(x^3)$$ and $$dv = x~dx$$ $$du = -\sin(x^3)3x^2~dx$$ and $$v=x^2/2$$ This will give us an even more complicated integral than we started with. So we need to think of something else.

Notice that the limits of integration are symmetric about $$x = 0$$. Let's see if the integrand is an odd function.

 $$f(x) = x \cos(x^3)$$ $$f(-x) = -x \cos(-x^3)$$ $$f(-x) = -x \cos(x^3)$$

Since $$f(-x) = -f(x)$$ the function is odd. When odd functions are integrated symmetrically about zero, the result is zero.

$$\displaystyle{ \int_{-1}^1{ x \cos(x^3) ~dx } = 0 }$$

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$$\displaystyle{ \int { x \exp(-x^2) ~dx } }$$

Problem Statement

$$\displaystyle{ \int { x \exp(-x^2) ~dx } }$$

Hint

$$\exp(x)$$ is another way to write $$e^x$$. So in this question $$\exp(-x^2) = e^{-x^2}$$.

Problem Statement

$$\displaystyle{ \int { x \exp(-x^2) ~dx } }$$

$$\displaystyle{ -\frac{1}{2} e^{-x^2} + C }$$

Problem Statement

$$\displaystyle{ \int { x \exp(-x^2) ~dx } }$$

Hint

$$\exp(x)$$ is another way to write $$e^x$$. So in this question $$\exp(-x^2) = e^{-x^2}$$.

Solution

 $$\displaystyle{ \int{ x e^{-x^2} ~dx } }$$ Try integration by substitution. $$u = -x^2$$ $$du = -2x~dx$$ $$\displaystyle{ \frac{du}{-2x} = dx }$$ $$\displaystyle{ \int{ xe^u \frac{du}{-2x} } }$$ $$\displaystyle{ -\frac{1}{2} \int{ e^u ~ du } }$$ $$\displaystyle{ -\frac{1}{2} e^u + C }$$ $$\displaystyle{ -\frac{1}{2} e^{-x^2} + C }$$

$$\displaystyle{ -\frac{1}{2} e^{-x^2} + C }$$

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$$\displaystyle{ \int{ \frac{1+4\ln x}{x} ~dx } }$$

Problem Statement

$$\displaystyle{ \int{ \frac{1+4\ln x}{x} ~dx } }$$

$$\ln x (1+2\ln x) + C$$

Problem Statement

$$\displaystyle{ \int{ \frac{1+4\ln x}{x} ~dx } }$$

Solution

 Try integration by substitution. $$u = \ln x$$ $$\displaystyle{ du = \frac{1}{x}dx }$$ $$x~du = dx$$ $$\displaystyle{ \int{ \frac{1+4u}{x} x ~du } }$$ $$\int{ 1 + 4u ~du }$$ $$\displaystyle{ u + \frac{4u^2}{2} + C }$$ $$\ln x + 2(\ln x)^2 + C$$ $$\ln x (1+2\ln x) + C$$

$$\ln x (1+2\ln x) + C$$

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Section 2

Calculate the derivative of $$h(x)=x^3 3^{x^2-1}$$.

Problem Statement

Calculate the derivative of $$h(x)=x^3 3^{x^2-1}$$.

$$h'(x) = x^2 3^{x^2-1} [ 3 + 2x^2 \ln 3 ]$$

Problem Statement

Calculate the derivative of $$h(x)=x^3 3^{x^2-1}$$.

Solution

 First, let's rewrite the $$3^{x^2-1}$$ term using the exponential function. On this site, we advocate this technique so that we don't have to memorize special cases. Let $$a = 3^{x^2-1}$$ Take the natural log of both sides of the equation. $$\ln a = \ln 3^{x^2-1}$$ $$\ln a = (x^2-1)\ln 3$$ Now reverse the process by setting the terms to the exponential function exponents. $$e^{\ln a} = e^{(x^2-1)\ln 3}$$ $$a = e^{(x^2-1)\ln 3}$$ So our function is now $$h(x)=x^3 e^{(x^2-1)\ln 3}$$. Now take the derivative of $$h(x)$$, starting with the product rule. $$\displaystyle{ h'(x) = x^3 \frac{d}{dx}\left[ e^{(x^2-1)\ln 3} \right] + e^{(x^2-1)\ln 3} \frac{d}{dx}[x^3] }$$ Now apply the chain rule to the derivative of the exponential term. $$\displaystyle{ h'(x) = x^3 e^{(x^2-1)\ln 3} \frac{d}{dx}[(x^2-1)\ln 3] + e^{(x^2-1)\ln 3} (3x^2) }$$ Before we go on, let's see if we can factor out any common terms. $$h'(x) = x^2 e^{(x^2-1)\ln 3} [ x \ln 3 (2x) + 3 ]$$ Although not absolutely necessary, replace the exponential term with the original term from the problem statement. $$h'(x) = x^2 3^{x^2-1} [ 3 + 2x^2 \ln 3 ]$$

$$h'(x) = x^2 3^{x^2-1} [ 3 + 2x^2 \ln 3 ]$$

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Calculate the derivative of $$\displaystyle{ y = \ln \left[ \frac{x\sqrt{x-3}}{(4x+5)^{10}} \right] }$$.

Problem Statement

Calculate the derivative of $$\displaystyle{ y = \ln \left[ \frac{x\sqrt{x-3}}{(4x+5)^{10}} \right] }$$.

$$\displaystyle{ y' = \frac{1}{x} + \frac{1}{2(x-3)} - \frac{40}{4x+5} }$$

Problem Statement

Calculate the derivative of $$\displaystyle{ y = \ln \left[ \frac{x\sqrt{x-3}}{(4x+5)^{10}} \right] }$$.

Solution

 $$\displaystyle{ y = \ln \left[ \frac{x\sqrt{x-3}}{(4x+5)^{10}} \right] }$$ $$y = \ln x + (1/2)\ln(x-3) - 10\ln(4x+5)$$ $$\displaystyle{ y' = \frac{1}{x} + \frac{1}{2(x-3)} - \frac{10}{4x+5}(4) }$$ $$\displaystyle{ y' = \frac{1}{x} + \frac{1}{2(x-3)} - \frac{40}{4x+5} }$$

$$\displaystyle{ y' = \frac{1}{x} + \frac{1}{2(x-3)} - \frac{40}{4x+5} }$$

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Calculate the derivative of the function $$y(x)$$ given for $$-3 \lt x \lt -1$$ by $$0 \lt y \lt \pi$$ and $$\cos y = x+2$$.

Problem Statement

Calculate the derivative of the function $$y(x)$$ given for $$-3 \lt x \lt -1$$ by $$0 \lt y \lt \pi$$ and $$\cos y = x+2$$.

$$\displaystyle{ \frac{dy}{dx} = -\frac{1}{\sin y} }$$

Problem Statement

Calculate the derivative of the function $$y(x)$$ given for $$-3 \lt x \lt -1$$ by $$0 \lt y \lt \pi$$ and $$\cos y = x+2$$.

Solution

 Use implicit differentiation. $$\displaystyle{ \frac{d}{dx}[\cos y] = \frac{d}{dx}[x+2] }$$ $$\displaystyle{ -\sin y \frac{dy}{dx} = 1 }$$ $$\displaystyle{ \frac{dy}{dx} = -\frac{1}{\sin y} }$$

Notes -
1. The limits on x and y allow our answer to make sense.
2. For $$-3 \lt x \lt -1$$, the original equation $$\cos y = x+2$$ is valid since the cosine function is bounded by -1 and 1.
3. For $$0 \lt y \lt \pi$$, the derivative makes sense and removes the possibility that the denominator is zero.
However, the limits on x and y do not affect how we solve the problem.

$$\displaystyle{ \frac{dy}{dx} = -\frac{1}{\sin y} }$$

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Find the solution of the differential equation $$\displaystyle{ \frac{dy}{dx} = x^3\sqrt{y} }$$ which satisfies $$y=2$$ when $$x=0$$.

Problem Statement

Find the solution of the differential equation $$\displaystyle{ \frac{dy}{dx} = x^3\sqrt{y} }$$ which satisfies $$y=2$$ when $$x=0$$.

$$\displaystyle{ y = \left[ \frac{x^4}{8} + \sqrt{2} \right]^2 }$$

Problem Statement

Find the solution of the differential equation $$\displaystyle{ \frac{dy}{dx} = x^3\sqrt{y} }$$ which satisfies $$y=2$$ when $$x=0$$.

Solution

 Use separation of variables technique. $$\displaystyle{ \frac{dy}{dx} = x^3\sqrt{y} }$$ $$\displaystyle{ \frac{dy}{y^{1/2}} = x^3 ~dx }$$ $$\displaystyle{ \int{ \frac{dy}{y^{1/2}} } = \int{ x^3 ~dx } }$$ $$\int{ y^{-1/2} ~ dy} = \int{ x^3 ~dx }$$ $$\displaystyle{ \frac{y^{1/2}}{1/2} = \frac{x^4}{4} + C }$$ Use the initial conditions to solve for C. $$2(\sqrt{2}) = 0 + C$$ $$2\sqrt{2} = C$$ $$\displaystyle{ y^{1/2} = \frac{x^4}{8} + \sqrt{2} }$$ Solve for y. $$\displaystyle{ y = \left[ \frac{x^4}{8} + \sqrt{2} \right]^2 }$$

Note - We need to be careful when solving for y. When we square both sides we could change the function. However, note that before we square both sides, the right side is always positive, so squaring it doesn't change the result.

$$\displaystyle{ y = \left[ \frac{x^4}{8} + \sqrt{2} \right]^2 }$$

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Part B - Questions 9-14

Instructions for Part B - - You may use your calculator. You have 40 minutes to complete this part of the exam. Show all your work. Correct answers are worth 1 point. The remaining points are earned by showing calculations and giving reasoning that justify your conclusions. Give exact, simplified answers. Each question is worth 10 points.

Consider the function $$f(x) = x \sqrt{4-x^2}$$.
(a) Find the domain of f.
(b) Find the intervals where f is increasing and the intervals where f is decreasing.
(c) Find the local maxima and minima of f.
(d) Sketch the graph of f.

Problem Statement

Consider the function $$f(x) = x \sqrt{4-x^2}$$.
(a) Find the domain of f.
(b) Find the intervals where f is increasing and the intervals where f is decreasing.
(c) Find the local maxima and minima of f.
(d) Sketch the graph of f.

Solution

(a) domain
We start with all real numbers. The only restriction is under the square root, which is required to be non-zero.

 $$4-x^2 \geq 0$$ $$4 \geq x^2$$ $$x^2 \leq 4$$ $$-2 \leq x \leq 2$$
 Final Answer - Part (a) domain: $$-2 \leq x \leq 2$$

(b) increasing and decreasing intervals
For starters, we need to calculate the derivative, set the result equal to zero and solve for critical values.

 First, use the product rule. $$\displaystyle{ f'(x) = x\frac{d}{dx}[(4-x^2)^{1/2}] + (4-x^2)^{1/2}(1) }$$ Apply the chain rule ot the derivative. $$\displaystyle{ f'(x) = x (1/2) (4-x^2)^{-1/2}(-2x) + (4-x^2)^{1/2} }$$ $$\displaystyle{ f'(x) = \left[ \frac{-x^2}{(4-x^2)^{1/2}} + (4-x^2)^{1/2} \right] }$$ Multiply the numerator and denominator by $$(4-x^2)^{1/2}$$ $$\displaystyle{ f'(x) = \frac{-x^2 + 4 - x^2}{(4-x^2)^{1/2}} }$$ $$\displaystyle{ f'(x) = \frac{4-2x^2}{(4-x^2)^1/2} }$$ Now we need to determine all values of $$x$$ where $$f'(x) = 0$$. This happens when the numerator is zero and the denominator is non-zero. $$4-2x^2 = 0$$ $$2x^2 = 4$$ $$x^2 = 2$$ $$x = \pm \sqrt{2}$$

The critical values are $$x = \pm \sqrt{2}$$. At these points, the denominator is non-zero. Now set up a table showing the intervals, test values and conclusions.

interval

$$(-2, -\sqrt{2})$$

test value x

$$x=-1.5$$

sign of $$f'(x)$$

$$f'(-1.5) \lt 0$$

conclusion

decreasing

interval

$$(-\sqrt{2}, \sqrt{2})$$

test value x

$$x=0$$

sign of $$f'(x)$$

$$f'(0) \gt 0$$

conclusion

increasing

interval

$$(-\sqrt{2}, 2)$$

test value x

$$x=1.5$$

sign of $$f'(x)$$

$$f'(1.5) \lt 0$$

conclusion

decreasing

 Final Answers - Part (b) $$(-2, -\sqrt{2})$$ decreasing $$(-\sqrt{2}, \sqrt{2})$$ increasing $$(-\sqrt{2}, 2)$$ decreasing

(c) local extrema
We found the critical points in the previous section as $$x=\pm\sqrt{2}$$.
So for $$x=-\sqrt{2}$$ the function goes from decreasing on the left to increasing on the right tells us that $$x=-\sqrt{2}$$ is a local minimum.
Similarly, for $$x=\sqrt{2}$$ the function goes from increasing on the left to decreasing on the right, telling us that $$x=\sqrt{2}$$ is a local maximum.

 Final Answers - Part (c) $$x=-\sqrt{2}$$ local minimum $$x=\sqrt{2}$$ local maximum

(d) plot

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We want to build a box that has a base of width w and length 2w. The materials for constructing the top and bottom of the box cost $14 per square foot and the materials for constructing the sides of the box cost$8 per square foot. The total volume of the box must be 60 cubic feet. Find the dimensions of the box that will minimize the cost of materials. Round your answer to two digits after the decimal point.

Problem Statement

We want to build a box that has a base of width w and length 2w. The materials for constructing the top and bottom of the box cost $14 per square foot and the materials for constructing the sides of the box cost$8 per square foot. The total volume of the box must be 60 cubic feet. Find the dimensions of the box that will minimize the cost of materials. Round your answer to two digits after the decimal point.

$$\text{width} \approx 2.34$$ ft., $$\text{length} = 2w \approx 4.69$$ ft, $$\text{height} \approx 5.47$$ ft.

Problem Statement

We want to build a box that has a base of width w and length 2w. The materials for constructing the top and bottom of the box cost $14 per square foot and the materials for constructing the sides of the box cost$8 per square foot. The total volume of the box must be 60 cubic feet. Find the dimensions of the box that will minimize the cost of materials. Round your answer to two digits after the decimal point.

Solution

The box has width w, length l and height h. This gives us an equation for volume as $$V=lwh$$. We are given that the length is 2w, so $$l=2w$$ and volume is now $$V=2w^2h$$. We are not told anything about the height but we are told that the volume needs to be 60 ft3. So from the volume equation was have $$60=2w^2h \to h = 30/w^2$$.
Now since all the dimensions are in terms of one variable, we can build the equation for cost.
We have a top and bottom with the same dimensions, so the area of each is $$A=lw \to A=2w^2$$. There are two of these areas, one for the top, one for the bottom and the cost of those is $$C_{t/b} = 2*A*14 = 2(2w^2)*14 = 56w^2$$.
For two of the sides, the area of each is $$A=w*h = w(30/w^2) = 30/w$$ and for the other two sides the area is $$A=2w*h = 2w(30/w^2) = 60/w$$. Each of those sides cost \$8 per square foot.
So we have an equation for total cost as $$C=56w^2 + 2*(30/w)*8 + 2*(60/w)*8$$. After some simplification, we get $$C = 56w^2+1440/w$$.
We want to minimize cost and since we have an equation for cost with only one variable, we can find $$dC/dw$$.

 $$C = 56w^2+1440w^{-1}$$ $$\displaystyle{ \frac{dC}{dw} = 56(2w) + 1440(-w^{-2}) }$$ $$\displaystyle{ 112w - \frac{1440}{w^2} = 0 }$$ $$\displaystyle{ 112w = \frac{1440}{w^2} }$$ $$\displaystyle{ w^3 = \frac{1440}{12} = \frac{90}{7} }$$ $$\displaystyle{ w = \left[ \frac{90}{7} \right]^{1/3} }$$

$$\text{width} \approx 2.34$$ ft., $$\text{length} = 2w \approx 4.69$$ ft, $$\text{height} \approx 5.47$$ ft.

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Lois Lane is in trouble and will die in 4 seconds. Superman is 100 miles away but Superman can fly at a speed given by $$v(t) = t^2 \sqrt{t^3+2}$$ (in miles per second). Will Superman be able to get there in time to rescue her?

Problem Statement

Lois Lane is in trouble and will die in 4 seconds. Superman is 100 miles away but Superman can fly at a speed given by $$v(t) = t^2 \sqrt{t^3+2}$$ (in miles per second). Will Superman be able to get there in time to rescue her?

Yes, he can reach her in time.

Problem Statement

Lois Lane is in trouble and will die in 4 seconds. Superman is 100 miles away but Superman can fly at a speed given by $$v(t) = t^2 \sqrt{t^3+2}$$ (in miles per second). Will Superman be able to get there in time to rescue her?

Solution

We are given the velocity which we need to use to find how long it will take to go 100 miles. A bit easier is to find out how far he can travel in 4 seconds. So we integrate the velocity function to get the distance function and then plug in $$t=4$$ seconds.

 $$s(t) = \int{ v(t) ~dt }$$ $$s(t) = \int{ t^2\sqrt{t^3+2} ~dt }$$ Use integration by substitution. $$u=t^3+2 \to du=3t^2~dt$$ $$\displaystyle{ \frac{1}{3}\int{ u^{1/2} ~du } }$$ $$\displaystyle{ \frac{1}{3}\frac{u^{3/2}}{3/2} + C }$$ $$\displaystyle{ s(t) = \frac{2}{9} (t^3+2)^{3/2} + C }$$ Although not given, use $$s(0)=0$$ to solve for $$C$$. $$\displaystyle{ 0 = \frac{2}{9}2^{3/2} + C }$$ $$\displaystyle{ C = -\frac{2}{9}2^{3/2} = -\frac{4}{9}\sqrt{2} }$$ So our equation for position is $$\displaystyle{ s(t) = \frac{2}{9} (t^3+2)^{3/2} - \frac{4}{9}\sqrt{2} }$$

Although we could solve this for $$t$$ to find the time to travel 100 miles, it is easier to plug in $$t=4$$ to see if he can go further than 100 miles in that time. Doing so, we get $$s(4)\approx 118.52$$ miles.
If you solve for $$t$$ instead, you should get something like $$[(450+2\sqrt{2})^{2/3}-2]^{1/3} \approx 3.85$$ seconds.

Yes, he can reach her in time.

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Consider the function$$f(x) = 4x^2 + \ln x$$, defined for $$x \gt 0$$.
(a) Verify that f has an inverse $$f^{-1}$$.
(b) Compute $$\left( f^{-1} \right)'(4)$$.

Problem Statement

Consider the function$$f(x) = 4x^2 + \ln x$$, defined for $$x \gt 0$$.
(a) Verify that f has an inverse $$f^{-1}$$.
(b) Compute $$\left( f^{-1} \right)'(4)$$.

(b) $$1/9$$

Problem Statement

Consider the function$$f(x) = 4x^2 + \ln x$$, defined for $$x \gt 0$$.
(a) Verify that f has an inverse $$f^{-1}$$.
(b) Compute $$\left( f^{-1} \right)'(4)$$.

Solution

(a) To verify that $$f(x)$$ has an inverse, we need to make sure it is one-to-one. Let's take the derivative.
$$f'(x) = 8x+1/x$$
Since the problem stated that x is positive, both terms in the derivative are positive and adding them also results in a positive value. So the derivative is always (for our problem) and therefore the function is always increasing and is one-to-one.

(b) To compute the derivative we need to know at which value of x, $$f(x)=4$$. This occurs at $$x=1$$. So the point $$(4,1)$$ is on the inverse function.
$$\begin{array}{rcl} \left( f^{-1} \right)'(4) & = & \displaystyle{ \frac{1}{f'(f^{-1}(4) )} } \\ & = & \displaystyle{ \frac{1}{f'(1)} } \\ & = & \displaystyle{ \frac{1}{8+1} } \\ & = & \displaystyle{ \frac{1}{9} } \end{array}$$

(b) $$1/9$$

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For the function given by $$\displaystyle{ G(x) = \int_{\sqrt{x}}^{2}{ \sin(t^2)~dt } }$$, find the derivative $$G'(x)$$.

Problem Statement

For the function given by $$\displaystyle{ G(x) = \int_{\sqrt{x}}^{2}{ \sin(t^2)~dt } }$$, find the derivative $$G'(x)$$.

$$G'(x) = \displaystyle{ \frac{-\sin(x)}{2\sqrt{x}} }$$

Problem Statement

For the function given by $$\displaystyle{ G(x) = \int_{\sqrt{x}}^{2}{ \sin(t^2)~dt } }$$, find the derivative $$G'(x)$$.

Solution

First, we need the variable in the upper limit. So we will rewrite the integral as $$\displaystyle{ G(x) = -\int_{2}^{\sqrt{x}}{ \sin(t^2)~dt } }$$.
To work this problem, we use the 2nd fundamental theorem of calculus which says that for $$\displaystyle{\frac{d}{dx} \left[ \int_{a}^{g(x)}{f(t)dt} \right]}$$, if we let $$u = g(x)$$, $$\displaystyle{ f'(x) = \frac{d}{du} \left[ \int_{a}^{u}{f(t)dt} \right] \frac{du}{dx}}$$.
So for this problem $$u = \sqrt{x}$$, so
$$\begin{array}{rcl} G'(x) & = & -\sin(u^2)(du/dx) \\ & = & -\sin(x)(1/2)x^{-1/2} \\ & = & \displaystyle{ \frac{-\sin(x)}{2\sqrt{x}} } \end{array}$$

$$G'(x) = \displaystyle{ \frac{-\sin(x)}{2\sqrt{x}} }$$

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A radioactive substance has a half-life of 1400 years. Suppose that initially there are 30 grams of the substance.
(a) Find an expression for the amount of the substance after t years.
(b) When will there be only 1 gram of the substance left? Round your answer to entire years.

Problem Statement

A radioactive substance has a half-life of 1400 years. Suppose that initially there are 30 grams of the substance.
(a) Find an expression for the amount of the substance after t years.
(b) When will there be only 1 gram of the substance left? Round your answer to entire years.

(a) $$C(t) = 30e^{-(t\ln 2)/1400}$$ grams     (b) 6870 years

Problem Statement

A radioactive substance has a half-life of 1400 years. Suppose that initially there are 30 grams of the substance.
(a) Find an expression for the amount of the substance after t years.
(b) When will there be only 1 gram of the substance left? Round your answer to entire years.

Solution

(a) This is an exponential growth/decay problem, so our general equation is $$C(t) = C_0 e^{-kt}$$. We are given that $$C(0) = 30$$ grams. So $$C_0 = 30$$. We are also told that the half-life is 1400 years. So half of the original amount is 15 grams. We use this information to solve for k.
$$\begin{array}{rcl} 15 & = & 30 e^{-k(1400)} \\ 0.5 & = & e^{-k(1400)} \\ \ln 0.5 & = & -1400k \\ \displaystyle{ -\frac{\ln 0.5}{1400} } & = & k \\ \displaystyle{ \frac{\ln 2}{1400} } & = & k \end{array}$$

 Final Answer - Part (a) $$C(t) = 30e^{-(t\ln 2)/1400}$$ grams

(b) We need to solve for t when $$C(t)=1$$ gram.
$$\begin{array}{rcl} 1 & = & 30 e^{-(t\ln 2)/1400} \\ \displaystyle{ \frac{1}{30} } & = & e^{-(t\ln 2)/1400} \\ -\ln 30 & = & \displaystyle{ \frac{-t\ln 2}{1400} } \\ t & = & \displaystyle{ \frac{1400\ln 30}{\ln 2} } \end{array}$$

 Final Answer - Part (b) $$\displaystyle{ t = \frac{1400\ln 30}{\ln 2} \approx 6870 }$$ years

(a) $$C(t) = 30e^{-(t\ln 2)/1400}$$ grams     (b) 6870 years

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### Topics You Need To Understand For This Page

 integrals separable differential equations graphing optimization velocity inverse functions and their derivatives exponential growth/decay all derivative topics

### Related Topics and Links

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### Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

Trig Derivatives

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

Inverse Trig Derivatives

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

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