Science:Math Exam Resources/Courses/MATH104/December 2010/Question 01 (k)

MATH104 December 2010

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Question 01 (k)
Find the global maximum and global minimum of the function $\displaystyle {}f(x)=x^{2/3}$ on the interval [-1,8].

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
What are the critical points? Is the interval of consideration closed?

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. On the closed interval $[-1,8]$, the global maximum/minimum can occur at critical points or at the endpoints. If $f(x)=x^{2/3}$, then $f'(x)={\frac {2}{3}}x^{-1/3}$ which is never $0$ but is undefined at $x=0$. The only critical point on the interval is at $x=0$. Now we evaluate $f$ at the critical point and at the endpoints: $f(0)=0$, $f(-1)=(-1)^{2/3}=1$, and $f(8)=8^{2/3}=4$. In comparing these we find the global maximum is $4$ occurring at $x=8$ and the global minimum is $0$ occurring at $x=0$.

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