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Key for: TEST 3

\begin{enumerate}

\item  F ($|x|$ is not differentiable on the interval).
F (this is only true if 2 is a critical point.  
$f(x)=-\sqrt{2}x^2$ has a local max at 0, but nowhere else.
If you check $f''(2)=\sqrt{2}$).
F (look at the picture on pg 189).
T (same picture).
F ($f(x)=h(x)+C$ but $C$ is not always 0)

\item 
\begin{enumerate}
\item  divide everything by $x^3$ this gives:
$=\lim_{x \rightarrow \infty} 
	\frac
	{4x^3-2x^{\frac{3}{2}}-1}
	{6x^3-x^{\frac{1}{2}}}$ 
$=\lim_{x \rightarrow \infty} 
	\frac
	{4-\frac{2}{x^{\frac{3}{2}}}-\frac{1}{x^3}}
		{6-\frac{1}{x^{\frac{5}{2}}}}$
$=\frac{4-0-0}{6-0}=\frac{4}{6}=\frac{2}{3}$

\item  Plugging in 3 gives $\frac{6}{0}$, so either $\infty$ or $-\infty$.
Checking sign shows that is must be $-\infty$.

\item $-1 \leq \sin x^2 \cos x^2 \leq 1$. 
Dividing this by $x^2$ now lets us use the squeeze theorem.(answer is 0).
Note: 
$\frac{\sin x^2 \cos x^2 }{x^2}
\neq \frac{\frac{\sin x^2}{x^2}\frac{\cos x^2 }{x^2}}{\frac{x^2}{x^2}}$
$=\frac{\sin x^2 \cos x^2 }{x^4}$
\end{enumerate}

\item $f'(x)=\frac{1}{2\sqrt{x}}$.
So $f$ is differentiable for $x>0$ so the interval $[4,36]$ is ok.
We need to find $c$ so that:
$f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{\sqrt{36}-\sqrt{4}}{36-4}=\frac{1}{8}$.
So we solve 
$\frac{1}{2\sqrt{x}}=\frac{1}{8}$ and get that $x=16$.
Note: if $\sqrt{x}=4$ then $x \neq 2$ but $x=16$.

\item  Want to maximize area of rectangle.  
So $A=(2x)y$, where $y=15-3x^2$.
So, $A=2x(15-3x^2)=30x-6x^3$.
Looking at the picture, the smallest that $x$ can be is 0, and the biggest
that $x$ can be is where the parabola intersects the $x$-axis. 
So we find the $x$ intercept of the parabola, solving $15-3x^2=0$.
This gives $x=\pm \sqrt{5}$.
We want $x\geq 0$ so we throw away the negative.
So the domain of $A$ is $x \in [0,\sqrt{5}]$.
Take the derivative of $A$, $A'=30-18x^2$.
Set this equal to 0 and solve:
$30-18x^2=0$,$x^2=\frac{30}{18}=\frac{5}{3}$.
So $x=\pm \sqrt{\frac{5}{3}}$.  
The negative is not in the domain, so throw it out.  
$x=\sqrt{\frac{5}{3}}$.
Does this give us maximum area??  We better check.
We have 3 critical points: 0, $\sqrt{\frac{5}{3}}$, $\sqrt{5}$.
(remember the endpoints).
Plug these into $A$ to see which gives the largest value.
$A(0)=0$,
$A(\sqrt{\frac{5}{3}})=20\sqrt{\frac{5}{3}}$,
$A(\sqrt{5})=0$.
So $x=\sqrt{\frac{5}{3}}$ gives the maximum.  
The $y$ associated with this is:
$y=15-3(\sqrt{\frac{5}{3}})^2=10$.
So the dimensions of the rectangle is $2\sqrt{\frac{5}{3}}$ by 10.
(remember the base was actually $2x$)

\item  
\begin{enumerate}
\item  $f$ is increasing on $(-\infty,-4) \bigcup (1,\infty)$
\item  $f$ is decreasing on $(-4,-2) \bigcup (-2,1)$
\item  Local maximums of $f$ at $x=-4$
\item  Local minimums of $f$ at $x=1$
\item  $f$ is concave up on $(-3,-2) \bigcup (0,2)$
\item  $f$ is concave down on $(-\infty,-3) \bigcup (-2,0) \bigcup (2,\infty)$
\item  Inflection points of $f$ at $x=-3,-2,0,2$
\end{enumerate}

\item  See page 185 for a graph and more details.
Domain of $f$ is all real numbers ($x^2+1=0$ has complex solutions).
So there are no vertical asymptotes.
To find the horizontal asymptote, take 
$\lim_{x \rightarrow \infty} \frac{x}{x^2+1}=0$.  
So $y=0$ is the horizontal asymptote.
$f'(x)=\frac{(1)(x^2+1)-x(2x)}{(x^2+1)^2}$
$=\frac{-x^2+1}{(x^2+1)^2}$ (be careful with the algebra!!!).
$f''(x)=\frac{2x(x^2-3)}{(x^2+1)^3}$ (be really careful with the algebra!!!).
This gives critical points of $-1,1$.
Possible inflections points of $-\sqrt{3},0,\sqrt{3}$.
Testing these shows the local max/mins and inflection points. 


\end{enumerate}



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