Answer to Question #286170 in Calculus for Abdullah

Question #286170

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





1
Expert's answer
2022-01-10T16:33:01-0500
f(x)=x2exf(x)=x^2e^{-x}


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

Find the fist derivative with respect to xx

f(x)=(x2ex)=2xexx2exf'(x)=(x^2e^{-x})'=2xe^{-x}-x^2e^{-x}

Find the critical number(s)


f(x)=0=>2xexx2ex=0f'(x)=0=>2xe^{-x}-x^2e^{-x}=0

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

x1=0,x2=2x_1=0, x_2=2

Critical numbers: 0,2.0, 2.

Find the second derivative with respect to xx

f(x)=(2xexx2ex)f''(x)=(2xe^{-x}-x^2e^{-x})'




=2ex2xex2xex+x2ex=2e^{-x}-2xe^{-x}-2xe^{-x}+x^2e^{-x}

=ex(24x+x2)=e^{-x}(2-4x+x^2)

f(0)=e0(24(0)+(0)2)=2>0f''(0)=e^{-0}(2-4(0)+(0)^2)=2>0

f(0)=e2(24(2)+(2)2)=2e2<0f''(0)=e^{-2}(2-4(2)+(2)^2)=-2e^{-2}<0

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



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

f(3)=(3)2e3=9e3f(3)=(3)^2e^{-3}=9e^{-3}

f(0)=(0)2e0=0f(0)=(0)^2e^{-0}=0

f(2)=(2)2e2=4e2f(2)=(2)^2e^{-2}=4e^{-2}

The function f(x)f(x) has a local maximum with value of 4e24e^{-2} at x=2.x=2.

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

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

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


ii)

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

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

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

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

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


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