Science:Math Exam Resources/Courses/MATH110/April 2017/Question 02 (b)

MATH110 April 2017

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Question 02 (b)
At what value of $x$ in the interval $[-2,2]$ does $f(x)=x^{2}e^{x}$ have its absolute maximum?

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
The absolute maximum of a differentiable function must occur either at a critical point (where $f'(x)=0$) or at one of the endpoints -2 or 2.

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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. In order to find the absolute maximum of the function $x^{2}e^{x}$ on $[-2,2]$, we first must locate the critical points of this function. That is, we must take the derivative and set it equal to zero. To take the derivative of $x^{2}e^{x}$, we need to use the product rule: $0=2xe^{x}+x^{2}e^{x}$ We factor out $e^{x}$ : $0=(2x+x^{2})e^{x}$ Observe that $e^{x}$ is never equal to zero. So the only solutions to this equation occur when $0=2x+x^{2}$ We factor x out of the equation: $0=x(2+x)$ which is zero when x = 0 or when x = -2. Now, we must plug in our critical points (0 and -2) into the function $x^{2}e^{x}$. Furthermore, we must plug in the endpoints (-2 and 2) of the domain. We get ${\begin{aligned}f(-2)&=(-2)^{2}e^{-2}=4e^{-2}\\f(0)&=(0)^{2}e^{0}=0\\f(2)&=2^{2}e^{2}=4e^{2}\end{aligned}}$ Of these 3 numbers, $4e^{2}$ is the largest. So f achieves its maximum value of $4e^{2}$ when $x=2$. Answer: $\color {blue}4e^{2}$