# How do you find the maximum and minimum values of the function f(x)= x - ((64x)/(x+4)) on the interval [0,13]?

Jul 13, 2018

There is only a local minimum at $\left(12 , - 36\right)$

#### Explanation:

Calculate the first derivative of $f \left(x\right)$

The derivative of

$\left(x\right) ' = 1$

And by the quotient rule

$\left(\frac{64 x}{x + 4}\right) ' = \frac{64 \left(x + 4\right) - 1 \left(64 x\right)}{x + 4} ^ 2$

$= \frac{256}{x + 4} ^ 2$

Therefore,

$f ' \left(x\right) = 1 - \frac{256}{x + 4} ^ 2 = \frac{\left({\left(x + 4\right)}^{2}\right) - 64}{x + 4} ^ 2$

$= \frac{{x}^{2} + 8 x + 16 - 256}{x + 4} ^ 2$

$= \frac{{x}^{2} + 8 x - 240}{x + 4} ^ 2$

$= \frac{\left(x + 20\right) \left(x - 12\right)}{x + 4} ^ 2$

The critical points are when

$f ' \left(x\right) = 0$

$\implies$, $\left\{\begin{matrix}x = - 20 \\ x = 12\end{matrix}\right.$

Build a variation chart in the interval $\left[0 , 13\right]$

$\textcolor{w h i t e}{a a a a}$$\text{Interval}$$\textcolor{w h i t e}{a a a a}$$\left(0 , 12\right)$$\textcolor{w h i t e}{a a a a}$$\left(12 , 13\right)$

$\textcolor{w h i t e}{a a a a}$$\text{sign f'(x)}$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$\text{ f(x)}$$\textcolor{w h i t e}{a a a a a a a a}$↘$\textcolor{w h i t e}{a a a a a a a}$↗

There is only a local minimum at $\left(12 , - 36\right)$

graph{x-((64x)/(x+4)) [-73.8, 113.7, -63, 30.8]}