 # Consider the function below. f(x)=x^{2}e^{-x} (a) Find the exact value of eozoischgc 2021-12-19 Answered
Consider the function below.
$f\left(x\right)={x}^{2}{e}^{-x}$
(a) Find the exact value of the minimum of f for $x\ge 0$.
Find the exact value of the maximum of f for $x\ge 0$.
(b) Find the exact value of x at which f increases most rapidly.
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Step 1
Given the function
$f\left(x\right)={x}^{2}{e}^{-x}$
Step 2
First find the critical points, the first derivative of the function to zero
$f\left(x\right)={x}^{2}{e}^{-x}$
${f}^{\prime }\left(x\right)={e}^{-x}\frac{d}{dx}{x}^{2}+{x}^{2}\frac{d}{dx}{e}^{-x}$
${f}^{\prime }\left(x\right)=2x{e}^{-x}-{x}^{2}{e}^{-x}$
${f}^{\prime }\left(x\right)=x{e}^{-x}\left(2-x\right)$
${f}^{\prime }\left(x\right)=0$
therefore,
$x{e}^{-x}\left(2-x\right)=0$
$x=0,x=2$
Step 3
To find the exact maximum or minimum, plug the value of critical point into the function
$f\left(0\right)=0$
$f\left(2\right)={\left(2\right)}^{2}{e}^{-2}=0.5413$
therefore,
the exact value of the minimum $f\left(x\right)=0$
the exact value of the maximum $f\left(x\right)=\frac{4}{{e}^{2}}$

We have step-by-step solutions for your answer! Karen Robbins
Step 4
(b)
To find the exact value of x at which f increases most rapidly.
${f}^{\prime }\left(x\right)={e}^{-x}\left(2x-{x}^{2}\right)$
$f{}^{″}\left(x\right)={e}^{-x}\frac{d}{dx}\left(2x-{x}^{2}\right)+\left(2x+{x}^{2}\right)\frac{d}{dx}\left({e}^{-x}\right)$
$f{}^{″}\left(x\right)={e}^{-x}\left(2-2x\right)+{e}^{-x}\left(2x-{x}^{2}\right)$
$f{}^{″}\left(x\right)={e}^{-x}\left(2-2x-2x+{x}^{2}\right)$
$f{}^{″}\left(x\right)={e}^{-x}\left({x}^{2}-4x+2\right)$
$f{}^{″}\left(x\right)=0$
$\left({x}^{2}-4x+2\right)=0$
$x=2+\sqrt{2},x=2-\sqrt{2}$
Plug these values into the first derivative and check whether the first derivative is positive and negative
${f}^{\prime }\left(2-\sqrt{2}\right)={e}^{-\left(2-\sqrt{2}\right)}\left[2\left(2-\sqrt{2}\right)-{\left(2-\sqrt{2}\right)}^{2}\right]$
${f}^{\prime }\left(2-\sqrt{2}\right)=0.461$
${f}^{\prime }\left(2-\sqrt{2}\right)>0$
${f}^{\prime }\left(2+\sqrt{2}\right)={e}^{-\left(2+\sqrt{2}\right)}\left[2\left(2+\sqrt{2}\right)-{\left(2+\sqrt{2}\right)}^{2}\right]$
${f}^{\prime }\left(2+\sqrt{2}\right)=-0.16$
${f}^{\prime }\left(2+\sqrt{2}\right)=<0$
therefore, the function is increasing the most rapidly at $x=\left(2-\sqrt{2}\right)$

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