Question

A. The function is strictly increasing.B. The function is strictly decreasing.C. The function is increasing and decreasing.D. The function is constant.

Answer 1

We have the function f(x):

$f(x)=2\sqrt[]{x+4}$

We can test if a function is strictly increasing if the first derivative f'(x) > 0 for the domain of f(x).

We then can derive f(x) and prove if f'(x) >0:

$\begin{gathered} f^{\prime}(x)=2\frac{d(\sqrt[]{x+4})}{dx} \\ f^{\prime}(x)=2\cdot\frac{1}{2}(x+4)^{-\frac{1}{2}} \\ f^{\prime}(x)=\frac{1}{\sqrt[]{x+4}} \end{gathered}$

The domain of f(x) is x ≥ -4. Then, for the domain of f(x), both the numerator and denominator are always positive.

Then, f'(x) will be positive for all the values of x of the domain.

Then, if f'(x) >0 for all x, we can conclude that f(x) is strictly increasing for all x in the domain.

Answer: A. The function is strictly increasing.