# What are the absolute extrema of f(x)=9x^(1/3)-3x in[0,5]?

Jun 4, 2018

The absolute maximum of $f \left(x\right)$ is $f \left(1\right) = 6$ and the absolute minimum is $f \left(0\right) = 0$.

#### Explanation:

To find the absolute extrema of a function, we need to find its critical points. These are the points of a function where its derivative is either zero or does not exist.

The derivative of the function is $f ' \left(x\right) = 3 {x}^{- \frac{2}{3}} - 3$. This function (the derivative) exists everywhere. Let's find where it is zero:

$0 = 3 {x}^{- \frac{2}{3}} - 3 \rightarrow 3 = 3 {x}^{- \frac{2}{3}} \rightarrow {x}^{- \frac{2}{3}} = 1 \rightarrow x = 1$

We also have to consider the endpoints of the function when looking for absolute extrema: so the three possibilities for extrema are f(1), f(0) and  f(5). Calculating these, we find that $f \left(1\right) = 6 , f \left(0\right) = 0 ,$ and $f \left(5\right) = 9 \sqrt[3]{5} - 15 \approx 0.3$, so $f \left(0\right) = 0$ is the minimum and $f \left(1\right) = 6$ is the max.