Science:Math Exam Resources/Courses/MATH100 A/December 2023/Question 27(c)/Solution 1

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.