# Find the exact location of all the relative and absolute extrema of the function f(x) = 40...

## Question:

Find the exact location of all the relative and absolute extrema of the function {eq}f(x) = 40 \sqrt x (x - 1); \quad x \geq 0 {/eq}

## Extreme Points

The relative extrema of a function {eq}\displaystyle y=f(x) {/eq} are the values of the function evaluated at the points {eq}\displaystyle c, {/eq} from the domain such that

{eq}\displaystyle f'(c)= 0, \text{ or } f'(c) - DNE {/eq} and the derivative changes signs at the point {eq}\displaystyle c. {/eq}

The absolute extrema are the maximum or minimum values of the function.

To find the absolute extreme, we obtain the maximum and minimum values of the function evaluated at the critical points and the end-points of the domain.

To find the extreme values, we need the critical points, which require the derivative function.

{eq}\displaystyle \begin{align*} f'(x)=\frac{d}{dx}\left(40\sqrt{x}\ (x-1)\right) =40\left(\frac{1}{2\sqrt{x}}(x-1)+\sqrt{x}\right)= \frac{x-1+2x}{2\sqrt{x}}=\frac{3x-1}{2\sqrt{x}} \end{align*} {/eq}

So, the critical points are given by {eq}\displaystyle f'(x)=0 \text{ or } f'(x) - DNE\\ \displaystyle 3x-1=0 \text{ or } x=0\\ \displaystyle x=\frac{1}{3}\in[0,\infty) \text{ or } x=0\in [0,\infty).\\ {/eq}

So, the critical points and the end-points of the domain are {eq}\displaystyle x=\frac{1}{3}, x=0. {/eq}

We will look at the sign of the derivative function,

{eq}\displaystyle \begin{array}{c|cccc} \text{critical points} & 0& & \frac{1}{3} & \\ \hline\\ x &0& \frac{1}{4} & \frac{1}{3} & 1\\ \hline\\ \text{ signs of }f'(x) & DNE & - &0 & + \end{array} {/eq}

Becasue the derivative changes from negative tot positive, the critical point {eq}\displaystyle \boxed{x=\frac{1}{3} \text{ is a local and global minimum point}} \text{ and the absolute minimum values is } f\left(\frac{1}{3}\right)=40\frac{1}{\sqrt{3}}\left(\frac{1}{3}-1\right)=\boxed{-\frac{80}{3\sqrt{3}}}. {/eq}

The critical point {eq}\displaystyle x=0, {/eq} is not an absolute maximum,

because the function in increasing monotonically after {eq}\displaystyle x=\frac{1}{3}, {/eq} as shown by the derivative sign table, therefore there is no relative or absolute maximum.