The Solution:
The derivative of the function f(x) is given as
[tex]\begin{gathered} f^{\prime}(x)=\frac{-2}{3x^{\frac{1}{3}}}\text{ } \\ \text{ where -}\inftyStep 1:
We shall find the critical value(s) of the function f(x).
Critical Value of the function f(x) is the value(s) of x for which the derivative f'(x) is equal to zero. That is, the value of x when
[tex]\begin{gathered} f^{\prime}(x)=\frac{-2}{3}x^{-\frac{1}{3}}=0 \\ \text{Solving for x, we get} \\ x=0 \\ So,\text{ the critical value of f(x) is 0} \end{gathered}[/tex]
Step 2:
We shall obtain the function f(x) by integrating the derivative f'(x) with respect to x.
[tex]\int f^{\prime}(x)dx=\int \frac{-2}{3x^{\frac{1}{3}}}dx=\int \frac{-2}{3}x^{-\frac{1}{3}}dx=-\frac{2}{3}\int x^{-\frac{1}{3}}dx[/tex][tex]\begin{gathered} -\frac{2}{3}\int x^{-\frac{1}{3}}dx=(\frac{-\frac{2}{3}x}{\frac{2}{3}}^{-\frac{1}{3}+1})+C=-x^{\frac{2}{3}}+C \\ \text{Where C is the constant of integration.} \\ \text{ So,} \\ f(x)=-x^{\frac{2}{3}}\text{ at the origin ( i.e at (0,0)} \end{gathered}[/tex]
Using Desmos graph plotter, we have the graph below:
Clearly from the graph above, we have that:
The function f(x) is increasing on the interval
[tex](-\infty,0)[/tex]
The function f(x) is decreasing on the interval
[tex](0,\infty)[/tex][tex]\begin{gathered} \text{ We shall test the relative extrema of f(x) by substituting 0 for x in f(x).} \\ f(x)=-x^{\frac{2}{3}} \\ f(0)=-(0)^{\frac{2}{3}}=0 \end{gathered}[/tex]
The relative extrema occur at (0,0) meaning the value of x at the point is 0, and it is a Global maximum type of relative extrema because f(x) has its greatest value at the point.
