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

Problem

ƒ(x)=x2*e(−x)

Solution

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

d(ƒ(x))/d(x)=d(x2)/d(x)*e(−x)+x2d(e(−x))/d(x)

d(ƒ(x))/d(x)=2*x*e(−x)+x2*(−e(−x))

d(ƒ(x))/d(x)=e(−x)*(2*x−x2)

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

e(−x)*(2*x−x2)=0

x*(2−x)=0

x=0,x=2

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

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

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

d2(ƒ(x))/(d(x)2)=e(−x)*(x2−4*x+2)

4. Evaluate the second derivative at the critical points to determine the nature of each point.

ƒ(0)″=e0*(0−4*(0)+2)=2

ƒ(2)″=e(−2)*(2−4*(2)+2)=−2*e(−2)

5. Determine extrema based on the sign of the second derivative.

Since *ƒ(0)″>0, there is a local minimum at *x=0

Since *ƒ(2)″<0, there is a local maximum at *x=2

6. Calculate the y-coordinates for the local extrema.

ƒ(0)=0*e0=0

ƒ(2)=2*e(−2)=4*e(−2)

Final Answer

Local Minimum: *(0,0), Local Maximum: *(2,4*e(−2))

Want more problems? Check here!