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\markright{Math 131, Midterm 2 Solutions}



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\begin{document}



\begin{center}



\textbf{College of the Holy Cross, Fall Semester, 2004} \\

\textbf{Math 131, Midterm 2 Solutions} \\

\end{center}





\begin{enumerate}

  

\item{} [5~points each]\quad Compute the following derivatives. You

  may use any correct method.



  \bd



  \item{(a)} $\ds\frac{d}{dx}\Bigl(5x\sqrt{x}-\frac{2}{x^3}+11x-4\Bigr)$



    \soln First rewrite $x\sqrt{x}=x^{3/2}$ and

    $\frac{2}{x^3}=2x^{-3}$. Then by the power rule,

    
    $$
    \frac{d}{dx}\left(5x^{3/2}-2x^{-3}+11x-4\right)

    =\frac{15}{2}x^{1/2}+6x^{-4}+11.

    $$



    \sep



  \item{(b)} $\ds\frac{d}{dt} (t^2e^{-5t})$



    \soln By the product rule,

        \[

        \frac{d}{dt} (t^2e^{-5t})=t^2\frac{d}{dt}(e^{-5t})+e^{-5t}\frac{d}{dt}(t^2)

        \]

    Using the chain rule on the first term, this equals

        \[

        t^2e^{-5t}(-5)+e^{-5t}(2t)

        \]

    which simplifies to $(2t-5t^2)e^{-5t}$. 

    \sep



  \item{(c)} $\ds\frac{d}{dz}8(z^2+4\cos z+2)^3$



    \soln By the chain rule,

        \[

        \frac{d}{dz}8(z^2+4\cos z+2)^3=24(z^2+4\cos z+2)^2(2z-4\sin z).

        \]



\vfil\eject



  \item{(d)} $\ds\frac{d}{dx}\Bigl(\frac{x}{\sin x}\Bigr)$



    \soln By the quotient rule,

        \[

        \frac{d}{dx}\Bigl(\frac{x}{\sin x}\Bigr)

        =\frac{\sin x-x\cos x}{\sin^2x}

        \]



  \ed



\sep



\item{} [15~points]\quad The graph of a function~$f$ is shown

  below with several points marked. Check the appropriate boxes.

  \begin{center}

    \input{midterm2a.eepic}

  \end{center}



  \[

  \begin{array}{|l||c|c|c|c|c|c|c|}

    \hline

    \text{Point at which} & A & B & C & D & E & F & G \\

    \hline\hline

    \text{$f$~is not continuous} &&\text{\checkmark}&\text{\checkmark}&&&&\text{\checkmark} \\

    \hline

    \text{$f$~is not differentiable} &\text{\checkmark}&\text{\checkmark}&\text{\checkmark}&&\text{\checkmark}&\text{\checkmark}&\text{\checkmark} \\

    \hline

    \end{array}

  \]



\vskip 0pt plus 0.2fil





\item{} [5~points each]\quad Compute the indicated limits. Show all

  work for full credit. 



  \bd



  \item{(a)} $\ds\lim_{x\to1}\frac{3x^2-5x-2}{x^2-4x+4}$



  \soln Since $\ds\lim_{x\to1}x^2-4x+4=1\neq0$,

    \[

    \lim_{x\to1}\frac{3x^2-5x-2}{x^2-4x+4}

    =\frac{\ds\lim_{x\to1}3x^2-5x-2}{\ds\lim_{x\to1}x^2-4x+4}

    =\frac{-4}{1}=-4

    \]



  \vfil\eject



  \item{(b)} $\ds\lim_{x\to2}\frac{3x^2-5x-2}{x^2-4x+4}$



  \soln Since $\ds\lim_{x\to2}3x^2-5x-2=0$ and

  $\ds\lim_{x\to2}x^2-4x+4=0$, we cannot just ``plug in'' to compute

  the limit. Factoring gives

  \[

  \frac{3x^2-5x-2}{x^2-4x+4}=\frac{(x-2)(3x+1)}{(x-2)^2}

  =\frac{3x+1}{x-2}\qquad\text{ for }x\neq2

  \]

  The resulting function $\frac{3x+1}{x-2}$ has a vertical asymptote

  at $x=2$ since $x=2$ is a root of the denominator and not a root

  of the numerator. Consequently

  \[

  \lim_{x\to2}\frac{3x^2-5x-2}{x^2-4x+4}=\lim_{x\to2}\frac{3x+1}{x-2}

  \]

  does not exist.

  

  \sep



  \item{(c)} $\ds\lim_{x\to1^-}3\cdot\frac{x-1}{|x-1|}+1$



  \soln For $x<1$, $x-1<0$ so $|x-1|=-(x-1)$ and therefore

  \[

  3\cdot\frac{x-1}{|x-1|}+1=3\cdot\frac{x-1}{-(x-1)}+1=-3+1=-2

  \]

  so

  \[

  \lim_{x\to1^-}3\cdot\frac{x-1}{|x-1|}+1=-2.

  \]



  \ed



  \sep



\item{} [5~points each]\quad Each part refers to the graph shown.

  \begin{center}

    \input{midterm2b.eepic}

  \end{center}



  \bd

  

  \item{(a)} Find all intervals on which~$f(x)>0$.



  \soln The graph lies above the $x$-axis between $B$ and $E$, so

  $f(x)>0$ for $B<x<E$.



  \vfil\eject



  \item{(b)} Find all intervals on which~$f'(x)>0$.



  \soln The function is increasing from $A$ to $D$, so $f'(x)>0$ for $A<x<D$.



  \sep

  \item{(c)} Find all intervals on which~$f''(x)>0$.



  \soln The graph is concave up from $B$ to $C$, so $f''(x)>0$ for $B<x<C$.



  \ed



  \sep



\item{} 



  \bd



  \item{(a)} [5~points]\quad State the limit definition of the

  derivative~$f'(x)$.



  \soln 

  \[

  f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}

  \]



  \sep

  \item{(b)} [10~points]\quad Use the \textbf{definition} to compute

  the derivative function of $f(x)=\ds\frac{1}{3x}$.



  \soln

  \begin{align*}

    f'(x)&=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\

    &=\lim_{h\to0}\frac{\ds\frac1{3(x+h)}-\frac1{3x}}{h}\\

    &=\lim_{h\to0}\frac{\frac{x-(x+h)}{3x(x+h)}}{h}\\

    &=\lim_{h\to0}\frac{-h}{3xh(x+h)}\\

    &=\lim_{h\to0}\frac{-1}{3x(x+h)}\\

    &=-\frac{1}{3x^2}

  \end{align*}



  \vfil\eject



  \item{(c)} [5~points]\quad Find the tangent line to the graph of

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



  \soln By part (b), the slope of the tangent line at $x=2$ is

  $f'(2)=-\frac1{3(2)^2}=-\frac1{12}$. Since $f(2)=\frac16$, the line

  passes through the point $(2,\frac16)$. Its equation is therefore

    \[

    y-\frac16=-\frac1{12}(x-2)

    \]

    by the point-slope formula.

  \ed



\sep



\item{} [15~points]\quad The world's population is about

  $P(t)=6e^{0.013t}$ billion people, with $t$~measured in years

  since~1999. Find~$P'(17)$. Write a sentence or two explaining the

  meaning of your answer; be sure to include a discussion of units.



  \soln By the chain rule

    \[

    P'(t)=6e^{0.013t}\cdot(0.013)

    \]

    so $P'(17)=6e^{0.013(17)}\cdot(0.013)\approx0.0973$. This

    means that in 2016 the world population will be growing at a rate

    of about 97.3 million people per year.



\end{enumerate}



\end{document}

