Find the Local Maxima and Minima f(x)=xe^(-x^2)

Problem

ƒ(x)=x*e^(−x^2)

Solution

1. Find the first derivative using the product rule and the chain rule.

d(ƒ(x))/d(x)=e^(−x^2)⋅d(x)/d(x)+x⋅d(e^(−x^2))/d(x)

d(ƒ(x))/d(x)=e^(−x^2)+x*(e^(−x^2)⋅(−2*x))

d(ƒ(x))/d(x)=e^(−x^2)−2*x^2*e^(−x^2)

d(ƒ(x))/d(x)=e^(−x^2)*(1−2*x^2)

2. Identify critical points by setting the first derivative equal to zero.

e^(−x^2)*(1−2*x^2)=0

1−2*x^2=0

x^2=1/2

x=±1/√(,2)

3. Find the second derivative to apply the Second Derivative Test.

d^2(ƒ(x))/(d(x)^2)=d()/d(x)*(e^(−x^2)−2*x^2*e^(−x^2))

d^2(ƒ(x))/(d(x)^2)=−2*x*e^(−x^2)−(4*x*e^(−x^2)−4*x^3*e^(−x^2))

d^2(ƒ(x))/(d(x)^2)=e^(−x^2)*(4*x^3−6*x)

4. Test the critical points using the second derivative.

At *x=1/√(,2),d^2(ƒ(x))/(d(x)^2)=e^(−1/2)*(4*(1/(2√(,2)))−6/√(,2))=e^(−1/2)*(2/√(,2)−6/√(,2))<0⇒Local Maxima

At *x=−1/√(,2),d^2(ƒ(x))/(d(x)^2)=e^(−1/2)*(4*(−1/(2√(,2)))+6/√(,2))=e^(−1/2)*(−2/√(,2)+6/√(,2))>0⇒Local Minima

5. Calculate the function values at these points.

ƒ(1/√(,2))=1/√(,2)*e^(−1/2)=1/√(,2*e)

ƒ*(−1/√(,2))=−1/√(,2)*e^(−1/2)=−1/√(,2*e)

Final Answer

Local Maxima: *(1/√(,2),1/√(,2*e)), Local Minima: *(−1/√(,2),−1/√(,2*e))

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