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\leftline{\bf  Math 135 - section 01 -  Precalculus Diagnostic Quiz Solutions}
\leftline{\bf  September 6,  2019}
\vspace{.2in}

\begin{enumerate}

\item[1)] Find all real numbers $x$ satisfying $|2x - 6| = 8$.
\item[] {\it Solution:}  The equation with absolute values is the same as $2x - 6 = 8$
or $2x - 6 = -8$.  In the first case, $x = 7$; in the second case $x = -1$.  

\item[2)] Solve for $t$:  \ \ $3t^2 - 4t + 1 = 0$ (find all real number solutions). 
\item[] {\it Solution:}  It's easiest to factor:
$$3t^2 - 4t + 1 = (3t - 1)(t - 1) = 0$$
when $3t - 1 = 0$ or $t - 1 = 0$.  Thus $t = 1/3$, or $t = 1$.  

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

\item[] {\it Solution:}  We have
\begin{align*}
\left(u^{-6}v^2 \right)^3 \left(\frac{v^{-2}}{u^2}\right)^{-1} &=
(u^{-18} v^6) \cdot u^2 v^2\\
&=u^{-16} v^8.
\end{align*}
The correct answer is A.  

\item[4)] Find common factors and cancel to simplify: 
$\displaystyle \frac{12 x}{5x-10} \cdot \frac{x^2-4}{2x+4}$.

\item[] {\it Solution:} We have
\begin{align*}
\frac{12 x}{5x-10} \cdot \frac{x^2-4}{2x+4} &= \frac{12x}{5(x - 2)} \cdot \frac{(x-2)(x+2)}{2(x+2)}\\
&=\frac{6x}{5}.
\end{align*}
The correct answer is B.

\item[5)] If $f(x) = 5x^2-11$, find $f(a+1) - f(a)$ and simplify.

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

\item[6)] Let $f(x) = x^2 - 3 x$ and $g(x) = x + 1$.  Which function is equal to $f(g(x))$?

\item[] {\it Solution:}
$$f(g(x)) = (x+1)^2 - 3(x + 1) = x^2 + 2x + 1 - 3x - 3 = x^2 - x - 2.$$
This is function D.  
\end{enumerate}

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