Answer to Question #286170 in Calculus for Abdullah

Question

Q: Find the local and absolute extreme values of the function on the given interval. Also

specify the intervals where function is increasing or decreasing

𝑓(𝑥) = 𝑥²e^-x

Answer 1

$f(x)=x^2e^{-x}$

Domain: $(-\infin, \infin)$

Find the fist derivative with respect to $x$

$f'(x)=(x^2e^{-x})'=2xe^{-x}-x^2e^{-x}$

Find the critical number(s)

$f'(x)=0=>2xe^{-x}-x^2e^{-x}=0$

$xe^{-x}(2-x)=0$

$x_1=0, x_2=2$

Critical numbers: $0, 2.$

Find the second derivative with respect to $x$

$f''(x)=(2xe^{-x}-x^2e^{-x})'$

$=2e^{-x}-2xe^{-x}-2xe^{-x}+x^2e^{-x}$

$=e^{-x}(2-4x+x^2)$

$f''(0)=e^{-0}(2-4(0)+(0)^2)=2>0$

$f''(0)=e^{-2}(2-4(2)+(2)^2)=-2e^{-2}<0$

i) If $x\in [-1, 3]$

$f(-1)=(-1)^2e^{-(-1)}=e$

$f(3)=(3)^2e^{-3}=9e^{-3}$

$f(0)=(0)^2e^{-0}=0$

$f(2)=(2)^2e^{-2}=4e^{-2}$

The function $f(x)$ has a local maximum with value of $4e^{-2}$ at $x=2.$

The function $f(x)$ has a local minimum with value of at $x=0.$

The function $f(x)$ has the absolute maximum on $[-1, 3]$ with value of $e$ at $x=-1.$

The function $f(x)$ has the absolute minimum on $[-1, 3]$ with value of at $x=0.$

ii)

If $x<0, f'(x)<0, f(x)$ decreases.

If $0<x<2, f'(x)>0, f(x)$ increases.

If $x>2, f'(x)<0, f(x)$ decreases.

The function $f(x)$ increases on $(0, 2).$

The function $f(x)$ decreases on $(-\infin, 0)\cup(2, \infin).$