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\leftline{\bf  Math 135 - section 01 -  Precalculus Diagnostic Quiz Solutions}
\leftline{\bf  August 30,  2013}


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\begin{enumerate}
\item[1.] Let $f(x) = x^2 - x$ and $g(x) = x + 1$.  Which function is $f(g(x))$?

A.  $x^2 - x$\ \ \ \ \ B. $x^2 + 3x$\ \ \ \ \ \ C. $x^2 + x$\ \ \ \ \ D. $x^2 + x + 2$

\item[] {\it Solution:}  $f(g(x)) = (x + 1)^2 - (x + 1) = x^2 + 2x + 1 - x - 1 = x^2 + x$.  
This is C.  

\item[2.] Find common factors and cancel to simplify: $\displaystyle \frac{12x}{3x-6} \cdot \frac{x^2-4}{2x+4}$\vspace{.1in}

A. \ $\displaystyle \frac{x^2+12x-4}{5x-2}$ \ \ \ \ \ B. \ $\displaystyle 2x$ \ \ \ \ \ C. \ $\displaystyle -\frac{4}{3}(x-1)$ \ \ \ \ \ D. \ $\displaystyle \frac{12x^3-48x}{6x^2-12}$

\item[] {\it Solution:}  Since $x^2 - 4 = (x - 2)(x + 2)$, $2x + 4 = 2(x + 2)$, and $3x - 6 = 3(x - 2)$, 
after cancelling all common factors, what is left is $2x$.  The correct answer is B.

\item[3.]  Which is equal to $\displaystyle \left(u^{-5}v^2 \right)^3 \left(\frac{v^2}{u}\right)^{-1}$?\vspace{.1in}

A. \ $u^{-16}v^8$ \ \ \ \ \ B. \ $u^{-14}v^{4}$ \ \ \ \ \ C. \ $u^{-9}v^{3}$ \ \ \ \ \ D. \ $(uv)^{-7}$

\item {\it Solution:} $\left(u^{-5}v^2 \right)^3 = u^{-15} v^6$ and $\left(\frac{v^2}{u}\right)^{-1} = u v^{-2}$.
So adding exponents, we see this is B:  $u^{-14} v^4$.  


\item[4.] If $f(x) = 5x^2-11$, what is $f(a+1) - f(a)$? 

\item[] {\it Solution:}  $f(a+1) - f(a) = 5(a+1)^2 - 11 - (5a^2 - 11) = 10a + 5$

\item[5.] Find all values of $x$ satisfying $2(x-2) > 5$. 

\item[] {\it Solution:}  This is the same as $2x > 5 + 4 = 9$, so all $x > 9/2$ are
solutions. 

\item[6.] Solve for $x$:  \ \ $2x^2 - x - 6=0$ (find all solutions). 

\item[] {\it Solution:}  The quadratic equation factors as $(2x + 3)(x - 2) = 0$, so 
the solutions are $x = -\frac{3}{2}$ and $x = 2$.  


\end{enumerate}

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