Answer
a. $ f(g(x))=\sqrt {\frac{1}{x}+1}$
b. Domain: $(-\infty,-1] \cup (0,\infty)$,
c. Range: $[0,1) \cup (1,\infty)$
a. $ g(f(x))=\frac{1}{\sqrt {x+1}}$
b. Domain: $(-1,\infty)$,
c. Range: $(0,\infty)$
Work Step by Step
a. To calculate the composite function, we place the inside function into the "x" value of the outside function. Thus:
$$f\circ g=f(g(x))=\sqrt {\frac{1}{x}+1}$$
For $f\circ g$ to be defined in $\mathbb{R}$, the denominator of $\frac{1}{x}$ cannot be zero$$x \ne 0$$ and the expression $\frac{1}{x} + 1$ under the root cannot be negative $$\frac{1}{x} + 1 \geq 0$$
For $x \gt 0$ the inequality becomes $x \geq -1$
and the intersection of $x \gt 0$ with $x \geq -1$ gives us $x \gt 0$
For $x \lt 0$ the inequality becomes $x \leq -1$
and the intersection of $x \lt 0$ with $x \leq -1$ gives us $x \leq -1$
Therefore, the domain of $f \circ g$ is the union of $x \leq -1$ and $x \gt 0$
which can be written as $(-\infty,-1] \cup (0,\infty)$
The range should resemble the range of the square root function in $\mathbb{R}$ which is $(0, \infty)$. However, because $\frac{1}{x}$ can't be zero, the expression $\frac{1}{x} + 1$ under the root can also never equal exactly $1$. We have to exclude $\sqrt 1$ or just $1$ from the range.
Therefore, the range of $f \circ g$ is $[0,1) \cup (1,\infty)$
b. We compose the function as we did above, by replacing x in the outer function with the inner function: $$g\circ f=g(f(x))=\frac{1}{\sqrt {x+1}}$$
The value under the root cannot be negative. And in this case because the root is the denominator of a fraction, it also cannot be zero. Therefore, $x + 1 \gt 0$
or $x \gt -1$
The domain of $g \circ f$ is $(-1,\infty)$
The range should resemble the range of the $\frac{1}{x}$ function which is $(-\infty,0) \cup (0, \infty)$ however because $\sqrt {x + 1}$ is always positive, the resulting fraction can never be negative.
The range of $g \circ f$ is $(0,\infty)$
