What are the absolute extrema of  f(x)= xsqrt(4-x^2) -x in [-1,2]?

Feb 24, 2017

Relative max. at x = 1/2 sqrt(1/2(15-sqrt(33))
Relative min. at right endpoint: $\left(2 , - 2\right)$

Explanation:

Rewrite equation:
$f \left(x\right) = x \sqrt{4 - {x}^{2}} - x = x {\left(4 - {x}^{2}\right)}^{\frac{1}{2}} - x$

Find the first derivative using the product rule $\left(u v\right) ' = u \cdot v ' + v \cdot u '$ and Power rule: $\left({u}^{n}\right) ' = n {u}^{n - 1} u ' :$

Let $u = x$, $u ' = 1$
and $v = {\left(4 - {x}^{2}\right)}^{\frac{1}{2}}$, $v ' = \frac{1}{2} {\left(4 - {x}^{2}\right)}^{- \frac{1}{2}} \left(- 2 x\right) = \frac{- x}{\sqrt{4 - {x}^{2}}}$

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

Find a common denominator and simplify:

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

Find the critical numbers by setting $f \left(x\right) ' = 0$ and simplifying:
$4 - 2 {x}^{2} - \sqrt{4 - {x}^{2}} = 0$;
$4 - 2 {x}^{2} = \sqrt{4 - {x}^{2}}$
${\left(4 - 2 {x}^{2}\right)}^{2} = 4 - {x}^{2}$
$16 - 16 {x}^{2} + 4 {x}^{4} = 4 - {x}^{2}$
$4 {x}^{4} - 15 {x}^{2} + 12 = 0$

Let $z = {x}^{2}$: $4 {z}^{2} - 15 z + 12 = 0$
Use the quadratic equation to find values of $a$:

$a = \frac{15 \pm \sqrt{33}}{8}$

Since $x = \pm \sqrt{a} , x = \pm \sqrt{\frac{15 \pm \sqrt{33}}{8}} = \pm \sqrt{\frac{15 \pm \sqrt{33}}{2 \sqrt{4}}}$

So x = +-1/2 sqrt(1/2(15+-sqrt(33)); $x \approx \pm 1.0756 , \pm 1.6103$

Use the first derivative test to see which critical numbers are relative maximums and relative minimums.

Intervals based on endpoints:
$\left[- 1 , 1.0756\right) , \left(1.0756 , 1.61030\right) , \left(1.61030 , 2\right]$
f'(0) > 0; f'(1.5) < 0; f'(1.8) < 0

Relative max. at x = 1/2 sqrt(1/2(15-sqrt(33))
Relative min. at right endpoint: $\left(2 , - 2\right)$