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\centerline{Mathematics 135 -- Calculus 1}
\centerline{Precalculus Review Problems}
\centerline{August, 2019}
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\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, September 6.
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\item{1.}  Find all values of $x$ that satisfy the given inequality
or inequalities:
\medskip
\itemitem{a)}  $-4x \ge 20$   -- {\it Answer:}  all $x \le -5$ (dividing by $-4$ to solve for $x$
flips the inequality because $-4 < 0$)
\itemitem{b)} $x + 1 > 4$, or $x + 2 < -1$  -- {\it Answer:}  all $x > 3$ together
with all $x < -3$
(could also be written as a union of intervals:  $(-\infty,-3) \cup (3,+\infty)$)
\itemitem{c)}  $x + 3 > 1$ and $x - 2 < 1$ -- {\it Answer:}  The first
inequality says $x > -2$ and the second says $x < 3$, so this is all $x$ between
$-2$ and $3$, or $-2 < x < 3$ (could also be written as the interval $(-2,3)$.)
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\item{2.}  
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\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:}  ${y^8\over x^{10}}$
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\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:
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\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)
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\item{5.}  Solve for $x$:
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\itemitem{a.}  $x^2 + x - 12 = 0$  -- {\it Answer:}  $x = 3,-4$ (factor the quadratic as $(x - 3)(x + 4)$, or 
use the quadratic formula.)
\itemitem{b.}  $4x^3 - 2x^2 - 2x = 0$  -- {\it Answer:}  $x = 0,1,-1/2$ (factor as $2x(2x^2 - x - 1) = 2x(2x + 1)(x - 1)$
\itemitem{c.}  $8x^2 - 8x - 3 = 0$  -- {\it Answer:}  $x = {8 \pm \sqrt{64 + 96} \over 16} = {2\pm\sqrt{10}\over 4}$
(using the quadratic formula)
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\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!)
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\itemitem{b.}  ${58\over 3(3+t^2)} + {1\over 3}$
-- {\it Answer:} ${t^2+61\over 3(3+t^2)}$
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\itemitem{c.}  ${2x\over 2x - 1} - {3x \over 2x + 5}$
-- {\it Answer:} ${-2x^2+13x\over (2x-1)(2x+5)}$
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\itemitem{d.}  ${ 1 + {1\over x}\over 1 - {1\over x^2}}$
-- {\it Answer:} ${x\over x-1}$
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\itemitem{e.}  ${2x(x+1)^{-1/2}-(x+1)^{1/2}\over x^2}$
-- {\it Answer:} ${x-1\over x^2\sqrt{x+1}}$
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\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{a)} Let   $f(x) = x^3-2x^2+3$.  Simplify as far as possible:
${f(a+h) - f(a) \over h}$. -- {\it Answer:}
$$\eqalign{
{f(a+h) - f(a) \over h} &= 
{(a+h)^3 - 2(a+h)^2 + 3 - a^3 + 2a^2 - 3\over h}\cr
&= {a^3 + 3a^2h + 3ah^2 + h^3 - 2a^2 - 4ah - 2h^2 + 3 - a^3 + 2a^2 - 3\over h}\cr
&= {3a^2h + 3ah^2 + h^3 - 4ah - 2h^2\over h}\cr
&= 3a^2 + 3ah + h^2 - 4a - 2h\cr}
$$
\itemitem{b)}  Same question for 
$f(x) = {1\over \sqrt{x}}$.   -- {\it Answer:}
$$\eqalign{
{ {1\over\sqrt{a+h}} - {1\over\sqrt{a}} \over h} &=
{\sqrt{a} - \sqrt{a+h} \over h\sqrt{a}\sqrt{a+h} }\cr
&= {\sqrt{a} - \sqrt{a+h} \over h\sqrt{a}\sqrt{a+h}}\cdot {\sqrt{a}+\sqrt{a+h}\over \sqrt{a}+\sqrt{a+h}}\cr
&= {-1\over \sqrt{a} \sqrt{a+h} ({\sqrt{a} + \sqrt{a+h}})}\cr}$$
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\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 a possible answer, which you 
can see either by stopping after the second line above, or by using
the double angle formula $\sin(2A) = 2\sin(A)\cos(A)$ on the final result.)
\bye
