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SAMPLE EXAM 1
(First quarter calculus - Final exam)

Instructions: Read each question carefully. There are a total of 12 questions.

Answer each question clearly in the space provided. If you decide to change your answer, cross out your old answer and neatly write your new answer. Be sure to show your work where asked to do so, and read the questions carefully. You do not need to show work where not asked. Partial credit will be given only if your work is clearly presented and mostly correct. You do not need to simplify except where it is specified.

Notes, books and calculators are not to be used.





1. (10 points): If

\begin{displaymath}f(x) = \sqrt{9-x}\end{displaymath}

and

g(x) = x3 + 1

then
\fbox{\rule[-.1cm]{0cm}{.7cm} $f \circ g(x) = $\space \hspace{2in} }







\fbox{\rule[-.1cm]{0cm}{.8cm}$g^{-1} (x) = $\space \hspace{2in} }







\fbox{\rule[-.1cm]{0cm}{.8cm}$ (f \circ g) '(x) = $\space \hspace{2.in} }.





2. (15 points): Determine a so that f(x) is continuous at every point:

\begin{displaymath}f(x) = \left\{ \begin{array}{ll}
x^3 + a - 2 & \mbox{if \ $x...
...$ } \\
ax^2 & \mbox{if \ $2 < x $ } \\
\end{array} \right. \end{displaymath}




\fbox{\rule[-.1cm]{0cm}{.8cm} a = \hspace{2in} }




b. The value of e is about 2.718. Use the differential to estimate the value of e.9. Simplify your answer.





\fbox{\rule[-.1cm]{0cm}{.8cm} $ e^{.9} \approx$\space \hspace{2in} }
















3. (15 points): f(x) = x3 - 3x2 - 1 .
a. The critical points of f are : \fbox{\rule[-.1cm]{0cm}{.8cm} \hspace{2in} }.





b. The maximum value of f(x) on the interval [-3,2] is : \fbox{\rule[-.1cm]{0cm}{.8cm} \hspace{1in} }.







c. The minimum value of f(x) on the interval [-1,1] is : \fbox{\rule[-.1cm]{0cm}{.8cm} \hspace{1in} }.







d. The minimum value of f(x) on the interval [-3,2] is : \fbox{\rule[-.1cm]{0cm}{.8cm} \hspace{1in} }.







e. f(x) is concave up on: \fbox{\rule[-.1cm]{0cm}{.8cm} \hspace{1in} }.







f. f(x) has inflection points at: \fbox{\rule[-.1cm]{0cm}{.8cm} \hspace{1in} }.





4. (20 points) A ball is thrown into the air from the top of a building. Its height in feet after t seconds is given by:

h(t) = -16t2 + 16t + 96


a. When does the ball hit the ground? \fbox{\rule[-.1cm]{0cm}{.8cm} \hspace{2in} }


b. What is the ball's velocity when it hits the ground? \fbox{\rule[-.1cm]{0cm}{.8cm} \hspace{2in} }


c. How tall is the building? \fbox{\rule[-.1cm]{0cm}{.8cm} \hspace{2in} }


d. What is the greatest height that the ball reaches? \fbox{\rule[-.1cm]{0cm}{.8cm} \hspace{2in} }






5. (10 points) a. Suppose that x,y satisfy the equation

-7x2 + 48xy + 7y2 = 28 .

Find dy/dx at the point (0,-2).

\fbox{\rule[-.1cm]{0cm}{1.cm} \hspace{2in} }






b. Find a formula for dy/dx at the point (x,y).

\fbox{\rule[-.1cm]{0cm}{1.cm} \hspace{2in} }










6. (30 points): Find the derivatives [No work need be shown]
a.

f(x) = 2x x2

\fbox{\rule[-.1cm]{0cm}{1.cm} $f'(x) = $\space \hspace{3in} }







b.

\begin{displaymath}f(x) = (\tan x)^9 \end{displaymath}

\fbox{\rule[-.1cm]{0cm}{1.cm} $f'(x) = $\space \hspace{3in} }







c.

f(x) = 2x2 - x

, \fbox{\rule[-.1cm]{0cm}{1.cm} $f'(x) = $\space \hspace{3in} }







d.

\begin{displaymath}f(x) = \frac{\cos x} {\ln x} \end{displaymath}

\fbox{\rule[-.1cm]{0cm}{1.cm} $f'(x) = $\space \hspace{3in} }





e.

\begin{displaymath}f(x) = \frac{ \ln {x^5}}{5}\end{displaymath}

\fbox{\rule[-.1cm]{0cm}{1.cm} $f'(x) = $\space \hspace{3in} }





7. (30 points): Find the limits. Write $+ \infty$ or $ - \infty$ when appropriate. Write ``Does not exist" if the limit doesn't exist and neither of these apply. [No work need be shown]
a.

\begin{displaymath}\lim_{x \to 3} \frac{x^2 - 9 }{ x^2 - x } = \end{displaymath}

\fbox{\rule[-.1cm]{0cm}{.8cm} \hspace{2in} }







b.

\begin{displaymath}\lim_{x \to -3} \frac{x^2 - 9 }{x + 3 } = \end{displaymath}

\fbox{\rule[-.1cm]{0cm}{.8cm} \hspace{2in} }







c.

\begin{displaymath}\lim_{x \to 0} \frac{\sin x^2 }{ \sin ^2 2x }= \end{displaymath}

\fbox{\rule[-.1cm]{0cm}{.8cm} \hspace{2in} }







d.

\begin{displaymath}\lim_{x \to \infty} \frac{3x }{ \sqrt{2x^2 -3} }= \end{displaymath}

\fbox{\rule[-.1cm]{0cm}{.8cm} \hspace{2in} }







e.

\begin{displaymath}\lim_{x \to \infty} (1 + 5/x)^{ 3x } = \end{displaymath}

\fbox{\rule[-.1cm]{0cm}{.8cm} \hspace{2in} }







f.

\begin{displaymath}\lim_{x \to -\infty} \frac{1 + \cos x^3 }{x^2 } = \end{displaymath}

\fbox{\rule[-.1cm]{0cm}{.8cm} \hspace{2in} }





8. (10 points): a. Give the precise definition of what it means for a function f(x) to be continuous at the point x=x0.







b. Where is the function f(x) continuous?

\begin{displaymath}f(x) = \left\{
\begin{array}{ll}
\frac{x }{x^2 + 1 } & \mbox...
... }{x^2 - 1 } & \mbox{if \ $0 < x $} \ \\
\end{array} \right. \end{displaymath}

\fbox{\rule[-.1cm]{0cm}{1.cm} \hspace{4in} }







9. (15 points) a. State the precise definition of the derivative of a function f(x).









b. Use the definition of the derivative to compute f'(x) when

\begin{displaymath}f(x) = \sqrt {x + 4}.\end{displaymath}

[Show your work].











10. (20 points) A cylindrical can is to be made with height h and radius r. It is to hold a volume of 4000 cubic cm. The material for the top and bottom costs 2 cents per square centimeter, and the material for the side costs 1 cent per square centimeter. What dimensions should it have so as to minimize the cost of the materials?
\fbox{\rule[-.1cm]{0cm}{1.cm} r = \hspace{2in} }





\fbox{\rule[-.1cm]{0cm}{1.cm} h = \hspace{2in} }















11. (10 points) For the function

\begin{displaymath}f(x) = 2x^2 - \ln x \end{displaymath}


a. Find the equation of the tangent line to the graph of f(x) at (1,2).

\fbox{\rule[-.1cm]{0cm}{1.3cm} \hspace{4.in} }.







b. Prove that f(x) is equal to 3 somewhere in the interval [1,2].

Show work carefully.





12. (25 points) Multiple choice - Circle the correct response.



i)

\begin{displaymath}\lim_{x \to \pi^- } \frac{\cos x}{x} = \end{displaymath}



a) $ - \pi$ b) $1/\pi$ c) $-1/ \pi$ d) $ \infty$ e) Does not exist f) None of above




ii)

\begin{displaymath}\lim_{x \to -\infty } \frac{\vert 4x\vert}{2x-7} = \end{displaymath}




a) 2 b) - 2 c) $ \infty$ d) $ - \infty$ e) 0 f) None of above




iii)

\begin{displaymath}\lim_{x \to 3^-} \frac{ x - 3 }{ 1/x - 1/3} = \end{displaymath}



a) 0 b) -3 c) -9 d) $ - \infty$ e) $ \infty$ f) None of above




iv)

\begin{displaymath}\lim_{x \to 0} \frac{ \sqrt {x+4} -2 }{ x} = \end{displaymath}



a) 0 b) $ \infty$ c) 1/4 d) 4 e) 1 f) None of above




v)

\begin{displaymath}\lim_{x \to 0} (1-x)^{1/x} = \end{displaymath}



a) 0 b) $ \infty$ c) 1 d) -e e) 1/e f) None of above




vi)

\begin{displaymath}f(x) = \frac{1}{10} \ln ( \frac{5+x}{5-x} ). \ \ f'(x) = \end{displaymath}


a)

\begin{displaymath}\frac{1}{10}\frac{5+x}{5-x} \end{displaymath}

b)

\begin{displaymath}\frac{x}{5x^2 - 125}\end{displaymath}

c)

\begin{displaymath}\frac{1}{10}\frac{5-x}{5+x} \end{displaymath}

d)

\begin{displaymath}10 \frac{5-x}{5+x} \end{displaymath}

e)

\begin{displaymath}\frac{1}{10} \ln ( \frac{5-x}{5+x} )\end{displaymath}


f) None of above




viii)

\begin{displaymath}f(x) = 2^{-x^2}, \ \ \ f'(x) = \end{displaymath}



a)

2-x2 -1

b)

\begin{displaymath}2^{-x^2 -1} \ln 2\end{displaymath}

c)

\begin{displaymath}(-2x) 2^{-x^2 } \ln 2\end{displaymath}

d)

(-2x) 2-x2

e)

(2x) 2-x2

f) None of above




 
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Next: About this document ...
Joel Hass
1999-11-01