Find h(g(x)) g(x)=(x-1)^3 and h(x) = cube root of x+1

$g (x) = {(x - 1)}^{3}$ and $h (x) = \sqrt[3]{x} + 1$

Step 1

Set up the composite result function.

$h (g (x))$

Step 2

Evaluate $h ({(x - 1)}^{3})$ by substituting in the value of $g$ into $h$ .

$h ({(x - 1)}^{3}) = \sqrt[3]{{(x - 1)}^{3}} + 1$

Step 3

Remove parentheses.

$h ({(x - 1)}^{3}) = \sqrt[3]{{(x - 1)}^{3}} + 1$

Step 4

Pull terms out from under the radical, assuming real numbers.

$h ({(x - 1)}^{3}) = x - 1 + 1$

Step 5

Combine the opposite terms in $x - 1 + 1$ .

Add $- 1$ and $1$ .

$h ({(x - 1)}^{3}) = x + 0$

Add $x$ and $0$ .

$h ({(x - 1)}^{3}) = x$