f(x)=x2e−x
Domain: (−∞,∞)
Find the fist derivative with respect to x
f′(x)=(x2e−x)′=2xe−x−x2e−x Find the critical number(s)
f′(x)=0=>2xe−x−x2e−x=0
xe−x(2−x)=0
x1=0,x2=2Critical numbers: 0,2.
Find the second derivative with respect to x
f′′(x)=(2xe−x−x2e−x)′
=2e−x−2xe−x−2xe−x+x2e−x
=e−x(2−4x+x2)
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∈[−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−2The 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 (−∞,0)∪(2,∞).
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