CPM Homework Help : APCALC Problem 1-40

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1-40.

State the domain for each of the functions below.

$f(x) =\frac { x } { x ^ { 2 } + 1 }$
Hint (a):
$0$ is excluded from the denominator of a fraction because division by $0$ would make the fraction undefined. What value of $x$ , if any, would make the denominator $x^2 + 1 = 0$ ?

$g(x) =\frac { 1 } { x } - \frac { x } { x + 1 }$
Hint (b):
Consider the domain of each part of the subtraction problem separately.
$\frac{1}{x}$ has a domain of $x\not= 0$ while $\frac {x}{x+1}$ has a domain of $x\not=-1$ .
Since $g(x)$ is the difference between $\frac{1}{x}$ and $\frac {x}{x+1}$ , both excluded values must be excluded from $g(x)$ .

$h(x)=\sqrt{x^2-9}$
Hint (c):
Typically, a square root function has a domain of $x ≥ 0$ , but in this case, $x$ is squared... meaning both positive or negative values will yield positive outputs. However, do not neglect to consider the $− 9$ . What values of $x$ , both positive and negative, will make $x^2 − 9 > 0$ ?
Answer (c):
$x ≤ − 3 \cup x ≥ 3$

$k(x)=\frac{\log\left(x-3\right)}{\sqrt{x+4}}$
Hint (d):
Consider the domain of the numerator and the domain of the denominator separately.
1. What is the domain of $\log(x − 3)$ ?
1. What is the domain of $\sqrt{(x+4)}$ ?
1. What value is excluded from the denominator of all fractions?
1. Now combine these results to find the domain of $k(x)$ .
Answer (d):
1. $x > 3$
2. $x ≥ −4$
3. $x ≠ −4$
4. $x > 3$

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