# What are the absolute extrema of f(x)=x-sqrt(5x-2) in(2,5)?

Jul 13, 2018

There are no absolute extrema in the interval $\left(2 , 5\right)$

#### Explanation:

Given: $f \left(x\right) = x - \sqrt{5 x - 2} \in \left(2 , 5\right)$

To find absolute extrema we need to find the first derivative and perform the first derivative test to find any minimum or maximums and then find the $y$ values of the end points and compare them.

Find the first derivative:

$f \left(x\right) = x - {\left(5 x - 2\right)}^{\frac{1}{2}}$

$f ' \left(x\right) = 1 - \frac{1}{2} {\left(5 x - 2\right)}^{- \frac{1}{2}} \left(5\right)$

$f ' \left(x\right) = 1 - \frac{5}{2 \sqrt{5 x - 2}}$

Find critical value(s) $f ' \left(x\right) = 0$:

$1 - \frac{5}{2 \sqrt{5 x - 2}} = 0$

$1 = \frac{5}{2 \sqrt{5 x - 2}}$

$2 \sqrt{5 x - 2} = 5$

$\sqrt{5 x - 2} = \frac{5}{2}$

Square both sides: $5 x - 2 = \pm \frac{25}{4}$

Since the domain of the function is limited by the radical:

5x - 2 >= 0; " "x >= 2/5

We only need to look at the positive answer:

$5 x - 2 = + \frac{25}{4}$

$5 x = \frac{2}{1} \cdot \frac{4}{4} + \frac{25}{4} = \frac{33}{4}$

$x = \frac{33}{4} \cdot \frac{1}{5} = \frac{33}{20} \approx 1.65$

Since this critical point is $< 2$, we can ignore it.

This means the absolute extrema are at the endpoints, but the endpoints are not included in the interval.