\magnification=\magstep1
\centerline{Mathematics 135 -- Calculus 1}
\centerline{Precalculus Review Problems}
\centerline{August 28, 2013}
\bigskip
\noindent
The following problems will not be collected or graded.  They are
similar to questions on the first diagnostic quiz to be given in
class on Friday, August 30.
\bigskip
\item{1.}  Find all values of $x$ that satisfy the given inequality
or inequalities:
\medskip
\itemitem{a)}  $-4x \ge 20$   -- {\it Answer:}  all $x \le -5$.
\itemitem{b)} $x + 1 > 4$, or $x + 2 < -1$  -- {\it Answer:}  all $x > 4$ together
with all $x < -3$
(could also be written as a union of intervals:  $(-\infty,-3) \cup (4,+\infty)$)
\itemitem{c)}  $x + 3 > 1$ and $x - 2 < 1$ -- {\it Answer:}  all $x$ between
$-2$ and $3$, or $-2 < x < 3$
(could also be written as the interval $(-2,3)$.)
\bigskip
\item{2.}  
\medskip
\itemitem{a)} Rewrite using positive exponents only: ${x^{-1/3} \over x^{1/2}}$
-- {\it Answer:}  ${1\over x^{5/6}}$
\itemitem{b)} Simplify:  $(x^2y^{-3})(x^{-5}y^3)$ -- {\it Answer:}  $x^{-3} = {1\over x^3}$
\itemitem{c)} Simplify:  $\left({x^3\over -27y^{-6}}\right)^{-2/3}$
-- {\it Answer:}  ${ 9 \over x^2 y^4}$
\itemitem{d)} Simplify:  $\left({x^{-3}\over y^{-2}}\right)^2 
\left({y\over x}\right)^4$  -- {\it Answer:}  ${y^8\over x^{10}}$
\bigskip
\item{3.}  A salesperson's monthly commission is 15\% on all sales 
over \$12000.  If the goal is to make a commission of at least \$3000
per month, what monthly sales figure should he or she attain?
-- {\it Answer:} Call the monthly sales
figure $x$.  (Assuming no commission on the sales under 12000), 
we want $(.15)(x - 12000) \ge 3000$ so $x \ge 32000$.
\item{4.}  Factor:
\medskip
\itemitem{a)}  $7a^4  - 42a^2b^2 + 49 a^3 b$  -- {\it Answer:} $7a^2(a^2-6b^2+7ab)$
\itemitem{b)}  $xe^{-2x} - x^3e^{-x}$ -- {\it Answer:} $xe^{-x}(e^{-x} - x^2)$
\itemitem{c)}  $6ac+3bc-4ad-2bd$ -- {\it Answer:}  $(2a+b)(3c-2d)$
\itemitem{d)}  $3x^2 - 6x - 24$ -- {\it Answer:} $(3x+6)(x-4)$
\itemitem{e)}  $9x^2 - 16y^4$  -- {\it Answer:}  $(3x-4y^2)(3x+4y^2)$ (difference of squares)
\bigskip
\item{5.}  Solve for $x$:
\medskip
\itemitem{a.}  $x^2 + x - 12 = 0$  -- {\it Answer:}  $x = 3,-4$ (factor the quadratic)
\itemitem{b.}  $4x^3 + 2x^2 - 2x = 0$  -- {\it Answer:}  $x = 0,1,-1/2$ (factor)
\itemitem{c.}  $8x^2 - 8x - 3 = 0$  -- {\it Answer:}  $x = {2\pm\sqrt{10}\over 4}$
(using the quadratic formula)
\bigskip
\item{6.}  Simplify:
\itemitem{a.}  ${2a^2 - 2b^2\over b - a}\cdot {4a+4b\over a^2 + 2ab + b^2}$
-- {\it Answer:} $-8$ (factor and cancel!)
\medskip
\itemitem{b.}  ${58\over 3(3+t^2)} + {1\over 3}$
-- {\it Answer:} ${t^2+61\over 3(3+t^2)}$
\medskip
\itemitem{c.}  ${2x\over 2x - 1} - {3x \over 2x + 5}$
-- {\it Answer:} ${-2x^2+13x\over (2x-1)(2x+5)}$
\medskip
\itemitem{d.}  ${ 1 + {1\over x}\over 1 - {1\over x^2}}$
-- {\it Answer:} ${x\over x-1}$
\medskip
\itemitem{e.}  ${2x(x+1)^{-1/2}-(x+1)^{1/2}\over x^2}$
-- {\it Answer:} ${x-1\over x^2\sqrt{x+1}}$
\bigskip
\item{7.}  Let $f(x) = (x+1)^3$ and $g(x) = x^2 - 1$.  
\itemitem{a)} What is the 
function $f(g(x))$?  -- {\it Answer:} $f(g(x)) = x^6$
\itemitem{b)} What is the function $g(f(x))$? -- {\it Answer:} $g(f(x)) = (x+1)^6 - 1$
\itemitem{c)} What is the function $f(x)g(x)$? -- {\it Answer:} $(x+1)^3(x^2-1) = (x+1)^4(x-1)$
\itemitem{d)} What is the domain of the function ${f(x)\over g(x)}$? -- {\it Answer:} all real numbers 
$x \ne -1, 1$. 
\item{8.} 
\itemitem{b)} Let   $f(x) = x^2-2x+3$.  Simplify as far as possible:
${f(a+h) - f(a) \over h}$. -- {\it Answer:}
$2a - 2 + h.$
\itemitem{b)}  Same question for 
$f(x) = {1\over \sqrt{x}}$.   -- {\it Answer:}
$$\eqalign{
{ {1\over\sqrt{a+h}} - {1\over\sqrt{a}} \over h} &=
{\sqrt{a} - \sqrt{a+h} \over h\sqrt{a}\sqrt{a+h} }\cr
&= {\sqrt{a} - \sqrt{a+h} \over h\sqrt{a}\sqrt{a+h}}\cdot {\sqrt{a}+\sqrt{a+h}\over \sqrt{a}+\sqrt{a+h}}\cr
&= {-1\over \sqrt{a} \sqrt{a+h} ({\sqrt{a} + \sqrt{a+h}})}\cr}$$
\bigskip
\item{9.}  Express in terms of the sine and cosine functions
and simplify:
$$g(x) = \csc^2(x) + \sec^2(x)$$
{\it Answer:}
$$\eqalign{
\csc^2(x) + \sec^2(x) &= {1\over \sin^2(x)} + {1\over \cos^2(x)}\cr
&= {\cos^2(x) + \sin^2(x) \over (\sin(x) \cos(x))^2}\cr
&= {1\over \left({1\over 2} \sin(2x)\right)^2}\cr
&= {4\over \sin^2(2x)}\cr}$$
(Note:  ${1\over (\sin(x) \cos(x))^2}$ is also an
acceptable answer.)
\bye
