Science:Math Exam Resources/Courses/MATH104/December 2011/Question 01 (e)/Solution 1

The absolute maximum of f may occur at the critical points (if it has any), or at the endpoints of its domain. We can solve this problem by finding the domain of f first, then by finding its critical points.

Determine the Domain of ƒ

Because of the square root, the domain of f is defined where $2-x^2$ is positive or zero. Therefore,

$\begin{array}{rl}0& \le 2-x^2\\ x^2& \le 2\\ x& \le \pm \sqrt{2}\end{array}$

The domain of the given function is

$[-\sqrt{2},+\sqrt{2}]$

Find the Critical Points of ƒ

The critical points of ƒ are located where its derivative is zero. Let's find the derivative of ƒ:

$\begin{array}{rl}\frac{df}{dx}& =\frac{d}{dx}\left(x\sqrt{2-x^2}\right)\\ & =(\frac{d}{dx}(x))\sqrt{2-x^2}+x(\frac{d}{dx}(\sqrt{2-x^2}))\\ & =\left(1\right)\sqrt{2-x^2}+x(\frac{1}{2\sqrt{2-x^2}}\frac{d}{dx}(2-x^2))\\ & =\sqrt{2-x^2}+x(\frac{1}{2\sqrt{2-x^2}}(-2x))\\ & =\sqrt{2-x^2}-\frac{2x^2}{2\sqrt{2-x^2}}\\ \end{array}$

Setting this to zero and solving for x yields the critical points:

$\begin{array}{rl}0& =\sqrt{2-x^2}-\frac{2x^2}{2\sqrt{2-x^2}}\\ \frac{2x^2}{2\sqrt{2-x^2}}& =\sqrt{2-x^2}\\ 2x^2& =2(2-x^2)\\ 2x^2& =4-2x^2\\ 4x^2& =4\\ x^2& =1\\ x& =\pm 1\end{array}$

Therefore, there are two critical points at -1 and +1.

Find Absolute Maximum on the Domain

The absolute maximum could occur at the two critical points, or at the endpoints of the domain.

$\begin{array}{rl}f(-\sqrt{2})& =0\\ f(-1)& =-1\\ f(+1)& =+1\\ f(+\sqrt{2})& =0\end{array}$

The absolute maximum occurs at x = +1.