Answer
(a) $$f(g(x))=\sqrt{4x-2}$$ $$[\frac{1}{2}, \infty)$$
(b) $$g(f(x))= 4\sqrt{x+1}-3$$ $$[-1, \infty)$$
(c) $$f(f(x))= \sqrt{\sqrt{x+1}+1}$$ $$[-1, \infty)$$
(d) $$g(g(x))= 16x-15$$ $$(-\infty, \infty)$$
Work Step by Step
$f(x)=\sqrt{x+1}$
$g(x)=4x-3$
(a)$ f(g(x))= \sqrt{4x-3+1} = \sqrt{4x-2}$
Since we have a square root in the function and we should have a non-negative number under the square root, the domain will be:
$4x-2\geq0$
$4x\geq2$
$x\geq \frac{1}{2}$
$[\frac{1}{2}, \infty)$
(b) $g(f(x))= 4\sqrt{x+1}-3$
We have the same restriction here:
$x+1\geq0$
$x\geq-1$
$[-1, \infty)$
(c) $f(f(x))= \sqrt{\sqrt{x+1}+1}$
This time we have something like a "double restriction":
$Domain= \left\{ \begin{array}{ll} x+1\geq0 \\ \sqrt{x+1}+1\geq0 \end{array} \right.
\left\{ \begin{array}{ll} x\geq-1 \\ \sqrt{x+1}\geq -1 \end{array} \right.$
The square root of a number is always greater than a negative number, so we can omit the second expression and say that the domain of this function is:
$[-1, \infty)$
(d) $g(g(x))= 4(4x-3)-3 = 16x-12-3 = 16x-15$
We have no restriction in this case, so the domain is:
$(-\infty, \infty)$
