Question:
Given
$$f(x)=\sqrt{36-x^2} \\
g(x)=\sqrt{x+1}$$
then find
$$\left(\frac fg\right)(x) = \sqrt{\frac{36-x^2}{x+1}}$$
and find its domain.
My Solution:
The domain comes from
$$x+1\ne0$$
and (intersection)
$$\frac{36-x^2}{x+1}\ge0.$$
The first part implies that
$x\ne-1$. But
$$\frac{36-x^2}{x+1} \ge 0$$
when $x$ is in $(-\infty,-6] \cup (-1,6].$
So my final answer is $(-\infty,-6] \cup (-1,6]$.
WebAssign's Answer: $(-1,6]$.
Best Answer
Because the new function is a quotient of two functions, the domains of the original functions need to be taken into account too. In this case $f(x)$ has domain $[-6,6]$ and $g(x)$ has domain $[-1,\infty)$, so these needs to be intersected with the calculated domain $(-\infty,-6]\cup(-1,6]$ to get the correct answer of $(-1,6]$.