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

MATH110 April 2017

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

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[show] 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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[show] 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: $\color {blue}4e^{2}$