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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 answers.
Part A  Questions 111
Instructions for Part A   You have 30 minutes to complete this part of the exam. No calculators are allowed. All questions in this section are worth 3 points, except for question 10, which is worth 4 points.
Find the indicated limit or determine that it doesn't exist.
Question 1 
\(\displaystyle{ \lim_{\theta \to 0}{\left[ \frac{\sin(\theta)}{\tan(3\theta)} \right]} }\) 


\(\displaystyle{ \lim_{\theta \to 0}{\left[ \frac{\sin(\theta)}{\tan(3\theta)} \right]} = 1/3}\) 
Direct substitution yields \( 0/0 \) which is indeterminate. So we need to do some algebra to be able to use one of the trig limits.
\(\displaystyle{
\begin{array}{rcl}
\lim_{\theta \to 0}{\left[ \frac{\sin(\theta)}{\tan(3\theta)} \right]} & = &
\lim_{\theta \to 0}{\left[ \sin(\theta) \frac{\cos(3\theta)}{\sin(3\theta)} \right]} \\
& = & \lim_{\theta \to 0}{\left[ \sin(\theta) \frac{\theta}{\theta} \frac{3}{3} \frac{\cos(3\theta)}{\sin(3\theta)} \right]} \\
& = & \lim_{\theta \to 0}{\left[ \frac{\sin(\theta)}{\theta} \frac{3\theta}{\sin(3\theta)} \frac{\cos(3\theta)}{3} \right]} \\
& = & \lim_{\theta \to 0}{\left[ \frac{\sin(\theta)}{\theta} \right]} \cdot \lim_{\theta \to 0}{\left[ \frac{3\theta}{\sin(3\theta)} \right]} \cdot \lim_{\theta \to 0}{\left[ \frac{\cos(3\theta)}{3} \right]} \\
& = & (1)(1)\frac{1}{3} = 1/3
\end{array}}\)
Question 1 Final Answer 
\(\displaystyle{ \lim_{\theta \to 0}{\left[ \frac{\sin(\theta)}{\tan(3\theta)} \right]} = 1/3}\) 
Question 2 
\(\displaystyle{ \lim_{x \to 2}{\left[ \frac{x^2+x6}{x2} \right]} }\) 


\(\displaystyle{ \lim_{x \to 2}{\left[ \frac{x^2+x6}{x2} \right]}=5}\) 
Direct substitution yields \( 0/0 \) which is indeterminate. Let's see if we can factor the numerator and, perhaps, cancel with the \( (x2) \) factor in the denominator.
\(\displaystyle{
\lim_{x \to 2}{\left[ \frac{x^2+x6}{x2} \right]} = }\) \(\displaystyle{
\lim_{x \to 2}{\left[ \frac{(x+3)(x2)}{(x2)} \right]} = }\) \(\displaystyle{ \lim_{x \to 2}{(x+3)} = 2+3 = 5 }\)
Question 2 Final Answer 
\(\displaystyle{ \lim_{x \to 2}{\left[ \frac{x^2+x6}{x2} \right]}=5}\) 
Question 3 
\(\displaystyle{ \lim_{x \to 3}{\left[ \frac{x^24x+6}{x^29} \right]} }\) 


\(\displaystyle{ \lim_{x \to 3}{\left[ \frac{x^24x+6}{x^29} \right]} }\) does not exist 
As usual, first try direct substitution.
\(\displaystyle{ \lim_{x \to 3}{\left[ \frac{x^24x+6}{x^29} \right]} = \frac{912+6}{99} = \frac{3}{0} }\)
So we have three possible situations here.
1. The limit goes to \( +\infty \).
2. The limit goes to \( \infty \).
3. The limit does not exist.
We need to look at what happens on both sides of \( x=3 \) to determine which case we have.
Left: \(\displaystyle{ \lim_{x \to 3^}{\left[ \frac{x^24x+6}{x^29} \right]} }\)
\( \to \) \(\displaystyle{ \frac{3}{negative} }\)
\( \to \) \(\displaystyle{ \infty }\)
Right: \(\displaystyle{ \lim_{x \to 3^+}{\left[ \frac{x^24x+6}{x^29} \right]} }\)
\( \to \) \(\displaystyle{ \frac{3}{positive} }\)
\( \to \) \(\displaystyle{ +\infty }\)
Since the limit approaches \( \infty \) on the left and \( +\infty \) on the right, and the left and right limits are not equal, the limit does not exist.
Note   Check the discussion here to understand the difference between when a limit goes to \( \pm \infty \) and when a limit does not exist.
Question 3 Final Answer 
\(\displaystyle{ \lim_{x \to 3}{\left[ \frac{x^24x+6}{x^29} \right]} }\) does not exist 
Question 4 
\(\displaystyle{ \lim_{x \to 1}{\left[ \frac{x1}{xx^3} \right]} }\) 


\(\displaystyle{ \lim_{x \to 1}{\left[ \frac{x1}{xx^3} \right]} = 1/2 }\) 
Direct substitution yields \( 0/0 \) which is indeterminate. So let's factor and see what we get.
\(\displaystyle{
\begin{array}{rcl}
\lim_{x \to 1}{\left[ \frac{x1}{xx^3} \right]} & = & \lim_{x \to 1}{\left[ \frac{x1}{x(1x^2)} \right]} \\
& = & \lim_{x \to 1}{\left[ \frac{x1}{x(x^21)} \right]} \\
& = & \lim_{x \to 1}{\left[ \frac{x1}{x(x1)(x+1)} \right]} \\
& = & \lim_{x \to 1}{\left[ \frac{1}{x(x+1)} \right]} \\
& = & \frac{1}{1(1+1)} = 1/2
\end{array}}\)
Question 4 Final Answer 
\(\displaystyle{ \lim_{x \to 1}{\left[ \frac{x1}{xx^3} \right]} = 1/2 }\) 
Question 5 
\(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{5x1}{3x^2+1/4} \right]} }\) 


\(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{5x1}{3x^2+1/4} \right]} = 0 }\) 
Direct substitution yields \( \infty / \infty \) which is indeterminate. Since we can't factor, let's try dividing the numerator and denominator by the highest factor, in this case \( x^2 \), and then use the Infinite Limits Theorem.
\(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{5x1}{3x^2+1/4} \frac{1/x^2}{1/x^2}\right]} = }\) \(\displaystyle{
\lim_{x \to \infty}{\left[ \frac{5/x1/x^2}{3+1/(4x^2)} \right]} = }\) \(\displaystyle{ \frac{0}{3} = 0 }\)
Question 5 Final Answer 
\(\displaystyle{ \lim_{x \to \infty}{\left[ \frac{5x1}{3x^2+1/4} \right]} = 0 }\) 
Evaluate the following derivatives and simplify. Give all exact answers.
Question 6 
\(\displaystyle{ \frac{d}{dx} \left[ (3x^2+x+1)^{2010} \right] }\) 


\(\displaystyle{ \frac{d}{dx} \left[ (3x^2+x+1)^{2010} \right] = 2010(6x+1)(3x^2+x+1)^{2009}}\) 
Use the chain rule.
\(\displaystyle{ \frac{d}{dx} \left[ (3x^2+x+1)^{2010} \right] = 2010(3x^2+x+1)^{2009} \cdot \frac{d}{dx} \left[ 3x^2+x+1 \right] = }\)
\(\displaystyle{ 2010(3x^2+x+1)^{2009} (6x+1) }\)
Question 6 Final Answer 
\(\displaystyle{ \frac{d}{dx} \left[ (3x^2+x+1)^{2010} \right] = 2010(6x+1)(3x^2+x+1)^{2009}}\) 
Question 7 
\(\displaystyle{ \frac{d}{dx}\left[ \frac{\sin^2 (x)}{\cos(x)} \right] }\) 


\(\displaystyle{ \frac{d}{dx}\left[ \frac{\sin^2 (x)}{\cos(x)} \right] = }\) \(\displaystyle{\frac{\sin(x)[ 2\cos^2 (x) + \sin^2 (x)]}{\cos^2 (x)}}\) 
Use the quotient rule.
\(\displaystyle{
\begin{array}{rcl}
\frac{d}{dx}\left[ \frac{\sin^2 (x)}{\cos(x)} \right] & = &
\frac{\cos(x)(2\sin(x))\cos(x)  \sin^2 (x)(\sin(x))}{\cos^2 (x)} \
& = & \frac{\sin(x)[ 2\cos^2 (x) + \sin^2 (x)]}{\cos^2 (x)}
\end{array}}\)
Question 7 Final Answer 
\(\displaystyle{ \frac{d}{dx}\left[ \frac{\sin^2 (x)}{\cos(x)} \right] = }\) \(\displaystyle{\frac{\sin(x)[ 2\cos^2 (x) + \sin^2 (x)]}{\cos^2 (x)}}\) 
Question 8 
Find \( D_x y\) for \( y = \sec^2 (4x) \) 


\( D_x [\sec^2 (4x)] = 8\sec^2 (4x)\tan(4x) \) 
Use the chain rule.
\(\displaystyle{
D_x [\sec^2 (4x) ] = 2 \sec(4x) \frac{d}{dx}[ \sec(4x) ] = }\) \(\displaystyle{
2\sec(4x)(\sec(4x)\tan(4x))(4) = }\) \(\displaystyle{ 8\sec^2 (4x)\tan(4x) }\)
Question 8 Final Answer 
\( D_x [\sec^2 (4x)] = 8\sec^2 (4x)\tan(4x) \) 
Question 9 
Find \(f'(3)\) for \( f(x) = (2+x^2)^{4/3}\) 


\( f'(3) = 8(11)^{1/3} \) 
Use the chain rule.
\(\displaystyle{ f'(x) = \frac{4}{3}(2+x^2)^{1/3}(2x) }\)
\(\displaystyle{ f'(3) = \frac{4}{3}(2+3^2)^{1/3}(6) = 8(11)^{1/3} }\)
Question 9 Final Answer 
\( f'(3) = 8(11)^{1/3} \) 
Question 10 
Find \( dy/dx\) using implicit differentiation for the curve \( 2xy^2 = \cos(xy)+5\) 


\(\displaystyle{ \frac{dy}{dx}=\frac{2+y \sin(xy)}{2yx \sin(xy)} }\) 
Use implicit differentiation.
\( \displaystyle{
\begin{array}{rcl}
\frac{d}{dx}\left[ 2xy^2 \right] & = & \frac{d}{dx} \left[ \cos(xy)+5 \right] \\
22y \frac{dy}{dx} & = & \frac{d}{dx} \left[ \cos(xy) \right] + 0 \\
& = & \sin(xy)\frac{d}{dx}[xy] \\
& = & \sin(xy) \left[ x\frac{dy}{dx} +y(1) \right] \\
& = & x \sin(xy) \frac{dy}{dx}  y \sin(xy) \\
2y\frac{dy}{dx} +x \sin(xy) \frac{dy}{dx} & = & 2 y \sin(xy) \\
(2y+x \sin(xy)) \frac{dy}{dx} & = & \\
\frac{dy}{dx} & = & \frac{2y \sin(xy)}{2y+x \sin(xy)}
\end{array}
}\)
Question 10 Final Answer 
\(\displaystyle{ \frac{dy}{dx}=\frac{2+y \sin(xy)}{2yx \sin(xy)} }\) 
Question 11 
Find \( D_x^{(101)} [ \cos(x) ]\) 


\( D_x^{(101)} [ \cos(x) ] = \sin(x)\) 
In order to work this problem, we need to find a pattern. So let's build a table.
\(
\begin{array}{lrcl}
0 & y & = & \cos(x) \\
1 & y' & = & \sin(x) \\
2 & y'' & = & \cos(x) \\
3 & y^{(3)} & = & \sin(x) \\
4 & y^{(4)} & = & \cos(x) \\
5 & y^{(5)} & = & \sin(x)
\end{array}
\)
So this repeats every 4th derivative.
\( D_x^{(101)} [ \cos(x) ] = D_x^1 [ \cos(x) ] = \sin(x)\)
Question 11 Final Answer 
\( D_x^{(101)} [ \cos(x) ] = \sin(x)\) 
Part B  Questions 1217
Instructions for Part B   You may use your calculator. You have one hour 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 answers.
Solve the following problems.
Question 12 
[12 points] Find the equation of the tangent line to \(\displaystyle{ y = \frac{1}{x^2+4} }\) at \( x=2\). Give your answer in slopeintercept form. 


\(\displaystyle{y = \frac{x}{16} + \frac{1}{4} }\) 
The equation of a line requires a point and a slope. To find the slope we need the derivative. We could use the quotient rule, however, since the numerator is a constant, it is easier to rewrite the function and use the chain rule.
\( y = (x^2+4)^{1} \)
\(\displaystyle{ y' = (1)(x^2+4)^{2} (2x) = \frac{2x}{(x^2+4)^2} }\)
Now we need the slope at \( x=2\).
\(\displaystyle{ y'(2) = \frac{2(2)}{(2^2+4)^2} = \frac{4}{8^2} = \frac{4}{64} = \frac{1}{16} }\)
So the slope at \(x=2\) is \( m=1/16\). Now we need the point. To find the yvalue of the function at \(x=2\), we calculate \( y(2)\).
\(\displaystyle{ y(2) = \frac{1}{2^2+4} = \frac{1}{8} }\)
So we have the point \((2,1/8)\) and slope \(m=1/16\). We can find the equation of the tangent line.
\(\displaystyle{
\begin{array}{rcl}
y  \frac{1}{8} & = & \frac{1}{16}(x  2) \\
y & = & \frac{x}{16} + \frac{1}{8} + \frac{1}{8} \\
y & = & \frac{x}{16} + \frac{1}{4}
\end{array}
}\)
Although we were not asked to graph the function and it's tangent line in the problem statement, if you have a graphing calculator with you, it's always a good idea to graph them quickly to make sure your answer makes sense.

produced by google 
Question 12 Final Answer 
\(\displaystyle{y = \frac{x}{16} + \frac{1}{4} }\) 
Question 13 
[12 points] Sketch the graph of a function \(g(x)\) that satisfies the following conditions. Be sure to label your graph and axes.
(a) \(g(x)\) is continuous at \(x=1\) (b) \(g'(1)\) does not exist
(c) \( g(1) = g(2) = 3 \) (d) \(\displaystyle{ \lim_{x \to 2^+}{g(x)} = \infty }\)
(e) \(\displaystyle{ \lim_{x \to \infty}{g(x)} = 1 }\) 

There are many possible answers. The key points are as follows.
(a) \(g(x)\) is continuous at \(x=1\)
This means that the limit from the right and left of \(x=1\) must be equal to the value of \(g(1)\).
(b) \(g'(1)\) does not exist
Together with part (a), this situation occurs when there is a sharp corner at that point.
(c) \( g(1) = g(2) = 3 \)
This gives the actual values we need to have at \(x=1\) and \(x=2\). To satisfy this requirement, we need to have two closed circles on the graph indicating these points. We also need to connect these points so as to satisfy requirement (a).
(d) \(\displaystyle{ \lim_{x \to 2^+}{g(x)} = \infty }\)
This requires that the righthanded limit at \(x=2\) go to negative infinity. So \(x=2\) is a vertical asymptote.
(e) \(\displaystyle{ \lim_{x \to \infty}{g(x)} = 1 }\)
Finally, this requires the limit of the function to go to \(1\) as \(x\) goes to \(+\infty\), i.e. we need a horizontal asymptote at \(y=1\). It doesn't matter if the limit is above or below the line \( y=1 \).
As long as your graph follows these requirements and it is a function (passes the vertical line test), then it will be correct. One possible solution is shown to the right.
Question 14 
[12 points] Identify all critical points and find the maximum and minimum value of the function
\( f(x) = x^33x^2+1 \) on \( S = [1,4]\). 


The critical points are \( (0,1)\) and \((2,3)\). The minimums occurs at the points \((1,3)\) and \((2,3)\) and the maximum occurs at the point \((4,17)\). 
Critical Points
Since \(f(x)\) is continuous for all real numbers, the only critical points will be where \( f'(x) = 0 \).
\( f'(x) = 3x^2  6x = 0 \)
\( 3x(x2)=0 \)
\( 3x = 0 ~~ \to ~~ x=0 \)
\( x2 = 0 ~~ \to ~~ x=2 \)
So, \( x=0\) and \(x=2\) are the critical numbers. Since the question used the term critical 'points', we need to find each yvalue for each corresponding critical number.
\( f(0) = 0^33(0)^2+1 = 1\)
\( f(2) = 2^3  3(2)^2+1 = 8 3(4)+1 = 812+1 = 3\)
So the critical points are \( (0,1) \) and \( (2,3) \).
Maximum and Minimum
Now we need to find the maximum and minimum values of the function. Since we have a closed interval, we need to check the endpoints as well.
Previously calculated, we have
\( f(0) = 1 \)
\( f(2) = 3 \)
\( f(1) = (1)^33(1)^2+1 = 13+1 = 3\)
\( f(4) = 4^3 3(4)^2_1 = 6448+1 = 17 \)
Comparing the numbers \( \{1, 3, 3, 17\} \), \(3\) is the smallest, so \( (1,3)\) and \((2,3)\) are the minimums; \(17\) is the largest, so \((4,17)\) is the maximum. Since this is a closed interval, these points are the absolute minimums and absolute maximum.
Although the problem statement did not ask for a graph, if you have a graphing calculator and some extra time after you have worked all the problems on the exam, it's a good idea to quickly plot a graph to make sure your answers make sense.
Question 14 Final Answer 
The critical points are \( (0,1)\) and \((2,3)\). The minimums occurs at the points \((1,3)\) and \((2,3)\) and the maximum occurs at the point \((4,17)\). 
Question 15 
[15 points] A ladder 3 meters long, with one end on the ground, is raised so that the angle \(\theta\) of the ladder with the ground increases by 2.5 radians per minute. Consider the right triangle formed by the ladder, the ground, and a vertical line through the top of the ladder. How fast is the area of this triangle increasing at the time when the angle \(\theta\) equals \(\pi/6\)? 


\(dA/dt = 45/8 ~ m^2/min \) 
First, let's draw the triangle. We have labeled the height \(h\) and base \(b\).
Given:
\(\displaystyle{ \frac{d\theta}{dt} = 2.5 rad/min }\) and ladder length \( 3 \) meters.
When \( \theta = \pi/6 \), we are asked to calculate \( dA/dt \) where \(A\) is the area of the triangle.
Equations:
\(
\begin{array}{rcl}
A & = & bh/2 \\
\cos(\theta) & = & b/3 ~~ \to ~~ b = 3\cos(\theta) \\
\sin(\theta) & = & h/3 ~~ \to ~~ h = 3\sin(\theta) \\
\end{array}
\)
Solving:
\(\displaystyle{ A = \frac{1}{2}(3\cos(\theta))(3\sin(\theta)) = \frac{9}{2}\cos(\theta)\sin(\theta) }\)
\(\displaystyle{
\frac{dA}{dt} = }\) \(\displaystyle{
\frac{9}{2} \left[ \cos(\theta)(\cos(\theta))\frac{d\theta}{dt} + \sin(\theta)(\sin(\theta))\frac{d\theta}{dt} \right ] = }\) \(\displaystyle{
\frac{9}{2}\left[ \cos^2(\theta)  \sin^2(\theta) \right] \frac{d\theta}{dt}
}\)
at \( \theta = \pi/6 ~~ d\theta/dt = 2.5 rad/min \)
\(\displaystyle{
\begin{array}{rcl}
\frac{dA}{dt} & = & \frac{9}{2} \left[ \cos^2(\pi/6)  \sin^2(\pi/6) \right] (m^2) (2.5 rad/min) \\
& = & \frac{9}{2} \left[ \frac{3}{4}  \frac{1}{4} \right] 2.5 m^2/min \\
& = & \left[ \frac{9}{4} \cdot \frac{5}{2} \right] m^2/min \\
& = & \frac{45}{8} m^2/min
\end{array}
}\)
Question 15 Final Answer 
\(dA/dt = 45/8 ~ m^2/min \) 

section 4  plot of f(x) 
For the questions in this section, consider the function \( f(x) = x^44x^310 \).
[ click here to show/hide the plot ]
Question 16 
[8 points] Find the intervals on which \(f(x)\) is decreasing and the intervals where \(f(x)\) is increasing. 


The function is decreasing on the intervals \((\infty,0)\) and \((0,3)\) and increasing on the interval \((3,\infty)\). 
Increasing and decreasing intervals are determined by the critical points. The critical points are points where the derivative is zero or the derivative doesn't exist. So our first step is to find the critical points.
\( f'(x) = 4x^3  12x^2 \)
\( 4x^3  12x^2 = 0 \)
\( 4x^2(x3) = 0 \)
\( 4x^2 = 0 ~~ \to ~~ x = 0 \)
\( x3=0 ~~ \to ~~ x=3 \)
So our critical values are \( x=0\) and \(x=3\). Now let's build a table to determine increasing and decreasing intervals.
Interval 
\( \infty < x < 0 \) 
\( 0 < x < 3 \) 
\( 3 < x < \infty \) 
Test xvalue 
\(x=1\)  \(x=1\)  \(x=4\) 
Sign of \(f'(x)\) 
\(f'(1)=16<0\)  \(f'(1)=8<0\)  \(f'(4)=64>0\) 
Conclusion 
decreasing  decreasing  increasing 
Although we were not asked to graph the function, it is always good to quickly plot it on your calculator to check your answers if there is adequate time.
Question 16 Final Answer 
The function is decreasing on the intervals \((\infty,0)\) and \((0,3)\) and increasing on the interval \((3,\infty)\). 
Question 17 
[7 points] Find the intervals on which \(f(x)\) is concave up and the intervals where \(f(x)\) is concave down. 


The function is concave up on the intervals \((\infty,0)\) and \((2,\infty)\) and concave down on the interval \((0,2)\). 
Change in concavity is determined by inflection points. Inflection points occur where the second derivative is zero or the second derivative does not exist. From the previous question, we know that \( f'(x) = 4x^3  12x^2 \).
\( f''(x) = 12x^224x \)
\( 12x^224x = 0 \)
\( 12x(x2) = 0 \)
\( 12x=0 ~~ \to ~~ x = 0\)
\( x2 = 0 ~~ \to ~~ x=2 \)
Since the second derivative is continuous for all real numbers, the only inflection points we have are \( x=0\) and \(x=2\).
So, now we will build a table to determine concavity.
Interval 
\( \infty < x < 0 \) 
\( 0 < x < 2 \) 
\( 2 < x < \infty \) 
Test xvalue 
\(x=1\)  \(x=1\)  \(x=3\) 
Sign of \(f''(x)\) 
\(f''(1)=36>0\)  \(f''(1)=12<0\)  \(f''(3)=36>0\) 
Conclusion 
concave up  concave down  concave up 
Although we were not asked to graph the function, it is always good to quickly plot it on your calculator to check your answers if there is adequate time.
Question 17 Final Answer 
The function is concave up on the intervals \((\infty,0)\) and \((2,\infty)\) and concave down on the interval \((0,2)\). 