# How do you determine the absolute extreme values for the function y=x(sqrt(1-x²)) on its domain?

Feb 14, 2018

The absolute minimum is:

$y \left(- \frac{1}{\sqrt{2}}\right) = - \frac{1}{2}$

and the absolute maximum is:

$y \left(\frac{1}{\sqrt{2}}\right) = \frac{1}{2}$

#### Explanation:

As a real function:

$y = x \sqrt{1 - {x}^{2}}$

is defined when $1 - {x}^{2} \ge 0$, that is for $x \in \left[- 1 , 1\right]$.
So the function is continuous on a compact domain and Weierstrass' theorem ensures it has an absolute maximum and an absolute minimum.

Evaluate first the value of $y$ and the limits of the domain:

$y \left(- 1\right) = y \left(1\right) = 0$

Evaluate now the first derivative:

$\frac{\mathrm{dy}}{\mathrm{dx}} = x \left(\frac{d}{\mathrm{dx}} \sqrt{1 - {x}^{2}}\right) + \left(\frac{d}{\mathrm{dx}} x\right) \sqrt{1 - {x}^{2}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = x \left(- \frac{x}{\sqrt{1 - {x}^{2}}}\right) + \sqrt{1 - {x}^{2}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \sqrt{1 - {x}^{2}} - {x}^{2} / \sqrt{1 - {x}^{2}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - {x}^{2} - {x}^{2}}{\sqrt{1 - {x}^{2}}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - 2 {x}^{2}}{\sqrt{1 - {x}^{2}}}$

and find the critical points of $y \left(x\right)$ inside its domain solving the equation:

$\frac{\mathrm{dy}}{\mathrm{dx}} = 0$

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

$1 - 2 {x}^{2} = 0$

$x = \pm \frac{1}{\sqrt{2}}$

Looking at the expression of $y ' \left(x\right)$ we can see that the denominator is always positive, while the numerator is a second degree polynomial with leading negative coefficient. $y ' \left(x\right)$ is therefore negative in the interval outside the roots and positive in the interval between the roots.

We can then conclude that:

$y \left(x\right)$ is decreasing in $\left(- 1 , - \frac{1}{\sqrt{2}}\right)$

$y \left(x\right)$ is increasing in $\left(- \frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}}\right)$

$y \left(x\right)$ is decreasing in $\left(\frac{1}{\sqrt{2}} , 1\right)$

so that $x = - \frac{1}{\sqrt{2}}$ is a local minimum and $x = \frac{1}{\sqrt{2}}$ is a local maximum. Evaluating:

$y \left(- \frac{1}{\sqrt{2}}\right) = \left(- \frac{1}{\sqrt{2}}\right) \sqrt{1 - \frac{1}{2}} = - \frac{1}{2}$

$y \left(\frac{1}{\sqrt{2}}\right) = \left(\frac{1}{\sqrt{2}}\right) \sqrt{1 - \frac{1}{2}} = \frac{1}{2}$

we can see that the minimum is lower and the maximum is higher than the values of $y \left(x\right)$ at the boundaries, so these are also the absolute minimum and maximum.