## Calculus: Early Transcendentals 8th Edition

(a) $(- \infty, -2) \cup (-2,1) \cup (1, \infty)$ (b) $(- \infty, \infty)$ (c) $(- \infty, -1] \cup [1,4]$
(a) $f(x) = \frac{2x+1}{x^2+x-2}$ $f(x) = \frac{2x+1}{(x+2)(x-1)}$ It is not admissible to have a value of zero in the denominator of the fraction. Therefore the domain includes all real numbers except $-2$ and $1$ We can express the domain as follows: $(- \infty, -2) \cup (-2,1) \cup (1, \infty)$ (b) $g(x) = \frac{\sqrt[3] x}{x^2+1}$ It is not admissible to have a value of zero in the denominator of the fraction. However, the denominator is positive for all values of $x$. Therefore, the domain includes all real numbers. We can express the domain as follows: $(- \infty, \infty)$ (c) $h(x) = \sqrt{4-x} - \sqrt{x^2-1}$ It is not admissible to have a negative value inside a square root. Therefore, $4-x \geq 0$ Then: $x \leq 4$ Also, $x^2-1 \geq 0$ Then: $x^2 \geq 1$ Thus: $x \leq -1$ or $x \geq 1$ We can express the domain as follows: $(- \infty, -1] \cup [1,4]$