Mathematics 131 – Calculus for Physical and Life Sciences 1 Precalculus Diagnostic Quiz – September 7, 2004

Solutions

1. Solve for x: 3x^{2} − 21x + 30 = 0.

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 ≠ 0, so the roots are given by

\begin{aligned}x − 2 = 0 & ⇒ x = 2, or \\ x − 5 = 0 & ⇒ x = 5. \end{aligned}

Method 2 (more calculation): Use the quadratic formula: ax^{2} + bx + c = 0 has roots

x = \frac{−b ± \sqrt{b^{2} − 4ac}}{2a}

Here a = 3, b = −21, c = 30, so

x = \frac{21 ± \sqrt{441 − 4(3)(30)}}{6} = \frac{21 ± \sqrt{81}}{6} = \frac{21 + 9}{6}, \frac{21 − 9}{6} = 5, 2.

Note: You could also factor out the 3, then apply the quadratic formula.

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))?

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} ≥ 0 is necessary. This is the same as x ≥ 0. The domain is the set of all x ≥ 0 in the real numbers.

3. Simplify and express as a single fraction:

\frac{\frac{x}{\sqrt{x+1}} + \sqrt{x + 1}}{\frac{x}{x^{2}+3}}

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:

\frac{\frac{x}{\sqrt{x+1}} + \sqrt{x + 1}}{\frac{x}{x^{2}+3}} = (\frac{x}{\sqrt{x + 1}} + \sqrt{x + 1}) · (\frac{x^{2} + 3}{x})

= (\frac{x}{\sqrt{x + 1}} + \frac{x + 1}{\sqrt{x + 1}}) · (\frac{x^{2} + 3}{x})

= (\frac{2x + 1}{\sqrt{x + 1}}) · (\frac{x^{2} + 3}{x})

= \frac{(2x + 1)(x^{2} + 3)}{x\sqrt{x + 1}}

(Many other forms possible too.)

4. Let f(x) = x^{2} + 2x. Compute and simplify: f(2 + h) − f(2 − h). 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:

[(2 + h)^{2} + 2(2 + h)] − [(2 − h)^{2} + 2(2 − h)] = [4 + 4h + h^{2} + 4 + 2h] − [4 − 4h + h^{2} + 4 − 2h]

= [8 + 6h + h^{2}] − [8 − 6h + h^{2}] = 12h