  ### Home > APCALC > Chapter 1 > Lesson 1.2.2 > Problem1-40

1-40.

State the domain for each of the functions below.

1. $f(x) =\frac { x } { x ^ { 2 } + 1 }$

$0$ 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 $x^2 + 1 = 0$?

1. $g(x) =\frac { 1 } { x } - \frac { x } { x + 1 }$

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 $g(x)$.

1. $h(x) =\frac { 1 } { x } - \frac { x } { x + 1 }$

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 $x^2 − 9 > 0$?

$x ≤ − 3 \cup x ≥ 3$

1. $k(x) =\frac { 1 } { x } - \frac { x } { x + 1 }$

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 $k(x)$.

1. $x > 3$

2. $x ≥ −4$

3. $x ≠ −4$

4. $x > 3$