We have the function f(x):
![f(x)=2\sqrt[]{x+4}](https://img.qammunity.org/2023/formulas/mathematics/college/xgxavf1y4qacid36al01at53y1nxpxuf81.png)
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(1)/(2)(x+4)^{-(1)/(2)} \\ f^(\prime)(x)=\frac{1}{\sqrt[]{x+4}} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/eytlwbyxwuh1umeg19yratdjhqzwo9vw5k.png)
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.