Question

a graphing calculator is recommended.
consider the following function. (if an answer does not exist, enter dne.)
$f(x)=x^{2} e^{-x}$
(a) find the exact value of the minimum of $f$ for $x geq 0$.
$f(x)=$
find the exact value of the maximum of $f$ for $x geq 0$.
$f(x)=$
(b) find the exact value of $x$ at which $f$ increases most rapidly.
$x=$

Explanation:

Step1: Find the first derivative

Use the product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u = x^{2}\), \(u^\prime=2x\), \(v = e^{-x}\), \(v^\prime=-e^{-x}\).
\(f^\prime(x)=2xe^{-x}-x^{2}e^{-x}=xe^{-x}(2 - x)\)
Set \(f^\prime(x)=0\), then \(x = 0\) or \(x = 2\) (since \(e^{-x}\gt0\) for all \(x\)).

Step2: Analyze the sign of \(f^\prime(x)\)

For \(0\lt x\lt2\), \(f^\prime(x)\gt0\), so \(f(x)\) is increasing.
For \(x\gt2\), \(f^\prime(x)\lt0\), so \(f(x)\) is decreasing.
At \(x = 0\), \(f(0)=0^{2}e^{-0}=0\).
At \(x = 2\), \(f(2)=4e^{-2}=\frac{4}{e^{2}}\).
As \(x\to+\infty\), \(\lim_{x
ightarrow+\infty}x^{2}e^{-x}=\lim_{x
ightarrow+\infty}\frac{x^{2}}{e^{x}}\), using L - H rule twice:
First application: \(\lim_{x
ightarrow+\infty}\frac{2x}{e^{x}}\), second application: \(\lim_{x
ightarrow+\infty}\frac{2}{e^{x}} = 0\).
So the minimum value of \(f(x)\) for \(x\geq0\) is \(0\) (at \(x = 0\)) and the maximum value is \(\frac{4}{e^{2}}\) (at \(x = 2\)).

Step3: Find the second derivative

\(f^\prime(x)=xe^{-x}(2 - x)=2xe^{-x}-x^{2}e^{-x}\)
\(f^{\prime\prime}(x)=2e^{-x}-2xe^{-x}-2xe^{-x}+x^{2}e^{-x}=e^{-x}(x^{2}-4x + 2)\)
Set \(f^{\prime\prime}(x)=0\), then \(x^{2}-4x + 2=0\).
Using the quadratic formula \(x=\frac{4\pm\sqrt{16 - 8}}{2}=2\pm\sqrt{2}\).
We want to find where \(f(x)\) increases most rapidly (i.e., where \(f^\prime(x)\) has a maximum).
Since \(x\geq0\), we consider \(x = 2-\sqrt{2}\) (because the function \(y = f^\prime(x)\) is a quadratic - like function (after factoring out \(e^{-x}\)) and we take the value in the domain \(x\geq0\)).

Answer:

(a) Minimum: \(0\), Maximum: \(\frac{4}{e^{2}}\)
(b) \(x = 2-\sqrt{2}\)