Answer
$$
f^{-1}(x)=x^{\frac{1}{4}}-2, \quad \quad x \geq 16.
$$
Work Step by Step
we first write
$$
y= f(x)=(x+2)^{4}, \quad x \geq 0
$$
Then we solve this equation for $ x$ as a function of $y$
$$
y^{\frac{1}{4}}= x+2
$$
$$
x =y^{\frac{1}{4}}-2
$$
which tells us that
$$
f^{-1}(y)=y^{\frac{1}{4}}-2. \quad \quad (i)
$$
Since we want $x$ to be the independent variable, we reverse $x$ and $y $ in (i) to produce the formula
$$
f^{-1}(x)=x^{\frac{1}{4}}-2.
$$
We know the domain of $f^{ −1}$ is the range of $f$
whereas the range of $ f(x)=(x+2)^{4}$ is $ [16, \infty) $. Thus, if we
want to make the domain of $f^{ −1}$ clear, we must express it explicitly by rewriting (i) as
$$
f^{-1}(x)=x^{\frac{1}{4}}-2, \quad x \geq 16.
$$
