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

MATH104 December 2011

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Question 01 (e)
Find the x-coordinate of the absolute maximum of $f(x)=x{\sqrt {2-x^{2}}}$ on its domain.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Find the domain of $f(x)$.

Hint 2
An absolute maximum will occur either at the endpoints of the domain or as a local maximum inside the domain.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

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, ${\begin{aligned}0&\leq 2-x^{2}\\x^{2}&\leq 2\\x&\leq \pm {\sqrt {2}}\end{aligned}}$ 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: ${\begin{aligned}{\frac {df}{dx}}&={\frac {d}{dx}}{\Big (}x{\sqrt {2-x^{2}}}{\Big )}\\&={\Big (}{\frac {d}{dx}}(x){\Big )}{\sqrt {2-x^{2}}}+x{\Big (}{\frac {d}{dx}}({\sqrt {2-x^{2}}}){\Big )}\\&={\Big (}1{\Big )}{\sqrt {2-x^{2}}}+x{\Big (}{\frac {1}{2{\sqrt {2-x^{2}}}}}{\frac {d}{dx}}(2-x^{2}){\Big )}\\&={\sqrt {2-x^{2}}}+x{\Big (}{\frac {1}{2{\sqrt {2-x^{2}}}}}(-2x){\Big )}\\&={\sqrt {2-x^{2}}}-{\frac {2x^{2}}{2{\sqrt {2-x^{2}}}}}\\\end{aligned}}$ Setting this to zero and solving for x yields the critical points: ${\begin{aligned}0&={\sqrt {2-x^{2}}}-{\frac {2x^{2}}{2{\sqrt {2-x^{2}}}}}\\{\frac {2x^{2}}{2{\sqrt {2-x^{2}}}}}&={\sqrt {2-x^{2}}}\\2x^{2}&=2(2-x^{2})\\2x^{2}&=4-2x^{2}\\4x^{2}&=4\\x^{2}&=1\\x&=\pm 1\end{aligned}}$ 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. ${\begin{aligned}f(-{\sqrt {2}})&=0\\f(-1)&=-1\\f(+1)&=+1\\f(+{\sqrt {2}})&=0\end{aligned}}$ The absolute maximum occurs at x = +1.

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