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Key for TEST 1

\begin{enumerate}

\item  To be continuous at a point $x$, a function $f$ must satisfy:
$\lim_{x\rightarrow c}f(x)=f(c)$
To be continuous, 
a function must continuous at all points in the domain of the function.
(A function must satisfy part all at all points in the domain)

\item Both are false.
For the first try $f(x)=x^2$ and $g(x)=x+1$.  
For these functions it is not true that $f\circ g=g\circ f$.(try it)
The second one is the condition that $f$ is continuous.
If this were true then all functions would be continuous, which is false.

\item  Solve for $y$ in the equation by getting the $y$'s on one side, 
factoring out a $y$ and then dividing.
$f(x)=y=\frac{3-\sin x}{2+x+x^2}$

\item  $f(x)=\sin x$ and $g(x)=x^2+\cos x$.
($f(x)=x$ and $g(x)=h(x)$ works so I told you not to use this one).

\item Either rationalize the denominator or factor the numerator:
$\frac{x-4}{\sqrt{x}-2}=\frac{(x-4)(\sqrt{x}+2)}{x-4}$ 
(rationalizing the denominator), or:
$\frac{x-4}{\sqrt{x}-2}=\frac{(\sqrt{x}-2)(\sqrt{x}+2)}{\sqrt{x}-2}$
(factoring the numerator)
Now, when you take the limit, you can cancel the factors and the limit is 4.

\item 1(the function value at -2 has no effect on the limit), 
2,-1, does not exist(since the left and right hand limits are not equal),
true, false.

\item 1,0, does not exist.
The best thing to do here is to graph it.
You should plot enough points to get a good idea of what is going on.
(I would plot $y$ values for $x=0,.1,.2,.3,.4,.5,.6,.7,.8,.9,1$)

\item Domain of $f$ is all real numbers except for -2.
If $f$ is going to be continuous at -2 then 
(from the definition of continuous), we need to define $f(-2)$ to be:
$\lim_{x\rightarrow -2}f(x)=\lim_{x\rightarrow -2}\frac{x^3+8}{x+2}
=\lim_{x\rightarrow -2}\frac{(x+2)(x^2-2x+4)}{x+2}
=\lim_{x\rightarrow -2}x^2-2x+4=12$.
So $f(-2)=12$ makes $f$ continuous.

\item  $a=7$ and $b=-11$. $f$ will be continuous except possibly at 2 and 3. 
(since $f$ is a line otherwise).
So we need to make sure the three lines connect up.
$\lim_{x\rightarrow 2^{-}}f(x)=\lim_{x\rightarrow 2^{-}}x+1=3$, and
$\lim_{x\rightarrow 2^{+}}f(x)=\lim_{x\rightarrow 2^{+}}ax+b=2a+b$
We need these to be equal, so 2a+b=3.
$\lim_{x\rightarrow 3^{-}}f(x)=\lim_{x\rightarrow 3^{-}}3x+1=10$, and
$\lim_{x\rightarrow 3^{+}}f(x)=\lim_{x\rightarrow 3^{+}}ax+b=3a+b$
We need these to be equal, so 3a+b=10.
You can now solve these two equations for $a$ and $b$.
(Another way to do this is to notice that $ax+b$ 
must be the equation of the line connecting (2,3) and (3,10))

\item  
$g'(x)=\lim_{h\rightarrow 0}\frac{g(x+h)-g(x)}{h}$
$=\lim_{h\rightarrow 0}\frac{\sqrt{2(x+h)-1}-\sqrt{2x-1}}{h}$
Now rationalize the numerator:

$=\lim_{h\rightarrow 0}
\frac{(\sqrt{2(x+h)-1}-\sqrt{2x-1})(\sqrt{2(x+h)-1}+\sqrt{2x-1})}
{h(\sqrt{2(x+h)-1}+\sqrt{2x-1})}$

$=\lim_{h\rightarrow 0}\frac{(2(x+h)-1-(2x-1)}{h(\sqrt{2(x+h)-1}+\sqrt{2x-1})}$

$=\lim_{h\rightarrow 0}\frac{2h}{h(\sqrt{2(x+h)-1}+\sqrt{2x-1})}$

$=\lim_{h\rightarrow 0}\frac{2}{(\sqrt{2(x+h)-1}+\sqrt{2x-1})}$
 $=\frac{2}{(\sqrt{2x-1}+\sqrt{2x-1})}$ 
$=\frac{1}{ \sqrt{2x-1}}$

To find the equation of the tangent line, the point on the graph is (5,3),
and the slope of the tangent line is $g'(5)=\frac{1}{3}$.
Now find the equation of the line through (5,3) with slope $\frac{1}{3}$
($y=\frac{1}{3}x-\frac{4}{3}$)



\end{enumerate}



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