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(x)=x2ex
(a) Find the exact value of the minimum of f for x0.
Find the exact value of the maximum of f for x0.
(b) Find the exact value of x at which f increases most rapidly.
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Expert Answer

Joseph Fair
Answered 2021-12-20 Author has 34 answers
Step 1
Given the function
f(x)=x2ex
Step 2
First find the critical points, the first derivative of the function to zero
f(x)=x2ex
f(x)=exddxx2+x2ddxex
f(x)=2xexx2ex
f(x)=xex(2x)
f(x)=0
therefore,
xex(2x)=0
x=0,x=2
Step 3
To find the exact maximum or minimum, plug the value of critical point into the function
f(0)=0
f(2)=(2)2e2=0.5413
therefore,
the exact value of the minimum f(x)=0
the exact value of the maximum f(x)=4e2

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Karen Robbins
Answered 2021-12-21 Author has 49 answers
Step 4
(b)
To find the exact value of x at which f increases most rapidly.
f(x)=ex(2xx2)
f(x)=exddx(2xx2)+(2x+x2)ddx(ex)
f(x)=ex(22x)+ex(2xx2)
f(x)=ex(22x2x+x2)
f(x)=ex(x24x+2)
f(x)=0
(x24x+2)=0
x=2+2,x=22
Plug these values into the first derivative and check whether the first derivative is positive and negative
f(22)=e(22)[2(22)(22)2]
f(22)=0.461
f(22)>0
f(2+2)=e(2+2)[2(2+2)(2+2)2]
f(2+2)=0.16
f(2+2)=<0
therefore, the function is increasing the most rapidly at x=(22)

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