Answer
(a) $(- \infty, -2) \cup (-2,1) \cup (1, \infty)$
(b) $(- \infty, \infty)$
(c) $(- \infty, -1] \cup [1,4]$
Work Step by Step
(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:
$(- \infty, -1] \cup [1,4]$
