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\leftline{Mathematics 131 -- Calculus for Physical and Life Sciences 1}
\leftline{Precalculus Diagnostic Quiz -- September 7, 2004}
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\leftline{Solutions}
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1.  Solve for $x$:  $3x^2 - 21x + 30 = 0$.
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Method 1 (easier):  Factoring, we see 
$3x^2 - 21x + 30 = 3(x - 2)(x - 5)$.  This equals zero only 
when one of the factors is zero.  $3 \ne 0$, so the roots are given by
$$\eqalign{x - 2 = 0 &\Rightarrow x = 2, {\rm\ or}\cr
           x - 5 = 0 &\Rightarrow x = 5.\cr}$$
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Method 2 (more calculation):  Use the quadratic formula:  $ax^2 + bx + c = 0$
has roots 
$$x = {-b \pm \sqrt{b^2 - 4ac} \over 2a}$$
Here $a = 3, b = -21, c = 30$, so 
$$x = {21 \pm \sqrt{441 - 4(3)(30)}\over 6} = {21 \pm \sqrt{81} \over 6} = 
{21 + 9\over 6}, {21 - 9\over 6} = 5,2.$$ 
Note: You could also factor out the 3, {\it then} apply the quadratic
formula.
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2.  Let $f(x) = \sqrt{x - 1}$ and $g(x) = x^3 + 1$.  Find $f(g(x))$?  
What is the domain of $f(g(x))$?
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$$f(g(x)) = \sqrt{g(x) - 1} = \sqrt{(x^3+1) - 1} = \sqrt{x^3} = x^{3/2}$$
The domain of a function defined this way is taken to be the biggest
set of real numbers you can substitute into the formula and get a 
well-defined result.   Here, we can't take the square root of 
a negative number and get a real value, so $x^3 \ge 0$ is necessary.
This is the same as $x \ge 0$.  The domain is the set of all $x\ge 0$
in the real numbers.
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3.  Simplify and express as a single fraction:
$${{x \over \sqrt{x+1}} + \sqrt{x+1} \over {x \over x^2 + 3}}$$
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First we rewrite the fraction as multiplying by the reciprocal of the 
denominator, then put the terms in the numerator over a common denominator
so they can be added,
then finally combine:
$$\eqalign{
{{x \over \sqrt{x+1}} + \sqrt{x+1} \over {x \over x^2 + 3}}
&= \left({x \over \sqrt{x+1}} + \sqrt{x+1}\right)\cdot \left({x^2 + 3 \over x}\right)\cr
&= \left({x \over \sqrt{x+1}} + {x+1\over \sqrt{x+1}}\right)\cdot \left({x^2 + 3 \over x}\right)\cr
&= \left({2x + 1\over \sqrt{x+1}}\right)\cdot \left({x^2 + 3 \over x}\right)\cr
&= {(2x+1)(x^2+3)\over x\sqrt{x+1}}\cr}$$
(Many other forms possible too.)
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4.  Let $f(x) = x^2 + 2x$.  Compute and simplify: $f(2+h) - f(2-h)$.
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In an expression like $f(2+h)$, we take the ``inside'' (the $2+h$)
and substitute that in for $x$ everywhere in the definition of the function:
$$f(2+h) - f(2-h) =
[(2+h)^2 + 2(2+h)] - [(2-h)^2 + 2(2-h)]$$
Now expand, collect like powers and simplify:
$$\eqalign{[(2+h)^2 + 2(2+h)] - [(2-h)^2 + 2(2-h)] &= [4+4h+h^2+4+2h]-[4-4h+h^2+4-2h]\cr
&=[8+6h+h^2]-[8-6h+h^2]\cr
&=12h\cr}$$

\bye