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\centerline{Mathematics 133 -- Intensive Calculus for Science 1}
\centerline{Precalculus Review Problems}
\centerline{August 28, 2001}
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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, August 31.
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\item{1.}  Find all values of $x$ that satisfy the given inequality
or inequalities:
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\itemitem{a)}  $-4x \ge 20$   -- {\it Answer:}  all $x \le -5$.
\itemitem{b)} $x + 1 > 4$, or $x + 2 < -1$  -- {\it Answer:}  all $x > 4$ or $x < -3$
(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 $x > -2$ and $x < 3$
(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 $x \ge 32000$.
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\item{4.}  The diameter $x$ in inches of a batch of ball bearings
manufactured by PAR Mfg. satisfies the inequality $|x - .1| \le .001$.
What are the largest and smallest diameters a ball bearing in 
the batch can have?  -- {\it Answer:}  $.099 \le x \le .101$.
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\item{5.}  Perform the indicated operations and simplify:
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\itemitem{a)}  $x - (2x - (-x - (1 - x)))$  -- {\it Answer:}  $-x - 1 = -(x+1)$
\itemitem{b)}  $2(t + \sqrt{t})^2 - 2t^2$ -- {\it Answer:} $2t(2\sqrt{t} + 1)$ (or $4t^{3/2} + 2t$)
\itemitem{c)}  $(2x^2-1)(x) - x^2(x + 2)$ -- {\it Answer:} $x^3 - 2x^2 - x$
\itemitem{d)}  $5x^2(3x+1)^4(6x) +(3x+1)^5(2x)$ -- {\it Answer:}  The 
``simplest'' form is probably the factored form: $2x(3x+1)^4(15x^2 + 2x + 1)$.
If you  expand and collect terms, you should get:
$2430 x^7 + 3726 x^6 + 2430 x^5 + 900 x^4 + 210x^3+30x^2 + 2x$ 
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\item{6.}  Factor out the greatest common 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:} $xe^{-x}(e^{-x} - x^2)$
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\item{7.}  Factor:
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\itemitem{a.}  $6ac+3bc-4ad-2bd$ -- {\it Answer:}  $(2a+b)(3c-2d)$
\itemitem{b.}  $3x^2 - 6x - 24$ -- {\it Answer:} $(3x+6)(x-4)$
\itemitem{c.}  $9x^2 - 16y^4$  -- {\it Answer:}  $(3x-4y^2)(3x+4y^2)$
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\item{8.}  Solve for $x$:
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\itemitem{a.}  $x^2 + x - 12 = 0$  -- {\it Answer:}  $x = 3,-4$
\itemitem{b.}  $4x^3 + 2x^2 - 2x = 0$  -- {\it Answer:}  $x = 0,1,-1/2$
\itemitem{c.}  $8x^2 - 8x - 3 = 0$  -- {\it Answer:}  $x = {2\pm\sqrt{10}\over 4}$
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\item{9.}  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{10.} 
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\itemitem{a.}  On a set of coordinate axes, plot the points $P = (1,3)$ and
$Q = (4,7)$.  Determine the distance $d(P,Q)$ between them.
-- {\it Answer:} distance is 5.
\itemitem{b.}  What is the Cartesian equation of the circle with
center $(h,k) = (4,2)$ and radius $r = 6$.
-- {\it Answer:} $(x-4)^2 + (y-2)^2 = 36$, or $x^2 - 8x + y^2 - 4y = 16$
\itemitem{c.}  The equation $x^2 + y^2 + x - 6y = 0$
defines a circle in the plane.  Find its center and radius.
-- {\it Answer:} Completing the square:  $(x+1/2)^2 + (y-3)^2 = 37/4$
center: $(-1/2,3)$, radius $\sqrt{37}/2$
\itemitem{d.}  Do the points $(3,4), (-3,7), (-6,1), (0,-2)$ form
the vertices of a square in the plane?  Why or why not?
-- {\it Answer:}  Yes -- Call the points 
$P = (3,4), Q = (-3,7), R = (-6,1), S = (0,-2)$.  Then the distance
from $P$ to $Q$ is 
$$\sqrt{(3+3)^2 + (4-7)^2} = \sqrt{45}= 3\sqrt{5}.$$
The distances from $Q$ to $R$, from $R$ to $S$, and $S$ to $P$ work
out to be the same, so $PQRS$ is a parallelogram with 4 equal sides (a
rhombus or ``diamond'').  Then, the line from $P$ to $Q$ has slope
${7 - 4\over -3 - 3} = {-1\over 2}$, while the line from $Q$ to 
$R$ has slope ${1-7\over -6 + 3} = 2$.  This shows that $PQ \perp QR$,
so the rhombus is a square.
\bye