Science:Math Exam Resources/Courses/MATH100 A/December 2023/Question 27(c)

MATH100 A December 2023

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Question 27(c)
Sketch the graph of $f$. Label all extrema and inflection points. Your label(s) should be at the point(s) in question, and coordinates should be written in the form $(-,-)$.

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
Inflection points occur at $x$-values for which $f''(x)=0$.

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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. First, the function itself passes through the origin. For $x>0$, $f(x)>0$. Second, we have computed that $f'(x)>0$ if $x\in (0,2)$, $f'(x)<0$ if $x>2$, and $f'(x)=0\iff x=2$. Thus, there is a local maximum at $x=2$. The point on the graph is $(2,f(2))=(2,16e^{-1})$. Third, we need to compute the second derivative of $f$ in order to find the inflection point(s) and concavity. We have, by the product rule, $f''(x)=-4\left({\frac {-1}{2}}e^{-x/2}(x-2)+e^{-x/2}\right)=2(e^{-x/2}(x-2)-2e^{-x/2})=2e^{-x/2}(x-4).$ Therefore, by a similar argument as in part (b), we see that $f''(x)=0\iff x=4$, and $f''(x)$ is positive (resp. negative) if and only if $x>4$ (resp. $0<x<4$). The inflection point is then $(4,f(4))=(4,32e^{-2})$. Finally, the picture should look something like the red graph in the adjacent figure. Graph of $f(x)=8xe^{-x/2}$, with extremum and inflection points indicated.