228k views
1 vote
A. The function is strictly increasing.B. The function is strictly decreasing.C. The function is increasing and decreasing.D. The function is constant.

A. The function is strictly increasing.B. The function is strictly decreasing.C. The-example-1
User Guig
by
5.1k points

1 Answer

2 votes

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(1)/(2)(x+4)^{-(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.

User Tounano
by
4.7k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.