Science:Math Exam Resources/Courses/MATH102/December 2013/Question C 03

MATH102 December 2013

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Question C 03
Consider the function $f(x)=x^{2}e^{-x}$ defined on the whole real line. Find all zeros of $\displaystyle f,f'$ and $\displaystyle f''$ and determine all local minima, maxima and inflection points of f. Sketch the graph of f. You may use the fact that $\lim _{x\to \infty }x^{2}e^{-x}=0$.

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
Take derivatives using the product rule. Then make your sign charts and plot the results.

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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. We begin by finding the zeroes of the original function, $\displaystyle 0=x^{2}e^{-x}$ and since $\displaystyle e^{-x}>0$ always, only $\displaystyle x=0$ is a root. But we note that $f$ is always greater or equal to zero. Next we take the derivative and see that $\displaystyle f'(x)=2xe^{-x}-x^{2}e^{-x}=xe^{-x}(2-x)$ Finding the roots gives us $\displaystyle x=0,2$. Making a sign chart, we see that The function is decreasing between $\displaystyle (-\infty ,0)$ The function is increasing between $\displaystyle (0,2)$ The function is decreasing between $\displaystyle (2,\infty )$ Graphically, we have and thus we have that $x=0$ is a local minimum and $x=2$ is a local maximum. The coordinates are given by $(0,0)$ and $(2,4e^{-2})$ (note that $4e^{-2}$ is roughly $4/9$). Lastly taking the second derivative, we have that $\displaystyle f''(x)=2(e^{-x}-xe^{-x})-(2xe^{-x}-x^{2}e^{-x})=e^{-x}(2-2x-2x+x^{2})=e^{-x}(2-4x+x^{2})$ The zeroes of the second derivative are given by $\displaystyle x^{2}-4x+2=0$. Using the quadratic formula gives us the roots $x={\frac {-(-4)\pm {\sqrt {(-4)^{2}-4(1)(2)}}}{2}}={\frac {4\pm {\sqrt {8}}}{2}}={\frac {4\pm 2{\sqrt {2}}}{2}}=2\pm {\sqrt {2}}$ Making a sign chart, we see that The function is concave up between $\displaystyle (-\infty ,2-{\sqrt {2}})$ The function is concave down between $\displaystyle (2-{\sqrt {2}},2+{\sqrt {2}})$ The function is concave up between $\displaystyle (2+{\sqrt {2}},\infty )$ Graphically we have Hence, we have that $2\pm {\sqrt {2}}$ are inflection points. Combining all this information, we have the following. First plot all known points of interest and label where the function is increasing, decreasing, concave up and concave down. Then connect the dots in a consistent way. Don't forget that $\lim _{x\to \infty }x^{2}e^{-x}=0$ was given to us and we can use this information.

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