# What are the absolute extreme values of the function f(x) = x/sqrt(x^2+1) on the interval [0,2]?

Aug 3, 2017

This function is strictly increasing on the given interval, so the absolute minimum value is $f \left(0\right) = 0$ and the absolute maximum value is $f \left(2\right) = \frac{2}{\sqrt{5}} \approx 0.894427$.

#### Explanation:

By the Quotient Rule and Chain Rule, the derivative of this function is $f ' \left(x\right) = \frac{\sqrt{{x}^{2} + 1} - x \cdot \left(\frac{1}{2}\right) {\left({x}^{2} + 1\right)}^{- \frac{1}{2}} \cdot 2 x}{{x}^{2} + 1}$.

This can be simplified by multiplying the top and bottom by $\sqrt{{x}^{2} + 1} = {\left({x}^{2} + 1\right)}^{\frac{1}{2}}$ to get

$f ' \left(x\right) = \frac{{x}^{2} + 1 - {x}^{2}}{{x}^{2} + 1} ^ \left(\frac{3}{2}\right) = \frac{1}{{x}^{2} + 1} ^ \left(\frac{3}{2}\right)$.

From this, we see that $f ' \left(x\right) > 0$ for all $x$. Therefore, $f$ is strictly increasing on any interval, including the interval $\left[0 , 2\right]$. This implies that the absolute minimum value is $f \left(0\right) = 0$ and the absolute maximum value is $f \left(2\right) = \frac{2}{\sqrt{5}} \approx 0.894427$.