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

Problem

ƒ(x)=x*e(−x2)

Solution

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

d(ƒ(x))/d(x)=e(−x2)⋅d(x)/d(x)+x⋅d(e(−x2))/d(x)

d(ƒ(x))/d(x)=e(−x2)+x*(e(−x2)⋅(−2*x))

d(ƒ(x))/d(x)=e(−x2)−2*x2*e(−x2)

d(ƒ(x))/d(x)=e(−x2)*(1−2*x2)

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

e(−x2)*(1−2*x2)=0

1−2*x2=0

x2=1/2

x=±1/√(,2)

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

d2(ƒ(x))/(d(x)2)=d()/d(x)*(e(−x2)−2*x2*e(−x2))

d2(ƒ(x))/(d(x)2)=−2*x*e(−x2)−(4*x*e(−x2)−4*x3*e(−x2))

d2(ƒ(x))/(d(x)2)=e(−x2)*(4*x3−6*x)

4. Test the critical points using the second derivative.

At *x=1/√(,2),d2(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),d2(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))

Want more problems? Check here!