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

Problem

ƒ(x)=x^2*e^(−x)

Solution

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

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

d(ƒ(x))/d(x)=2*x*e^(−x)+x^2*(−e^(−x))

d(ƒ(x))/d(x)=e^(−x)*(2*x−x^2)

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

e^(−x)*(2*x−x^2)=0

x*(2−x)=0

x=0,x=2

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

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

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

d^2(ƒ(x))/(d(x)^2)=e^(−x)*(x^2−4*x+2)

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

ƒ(0)^″=e^0*(0^2−4*(0)+2)=2

ƒ(2)^″=e^(−2)*(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)=0^2*e^0=0

ƒ(2)=2^2*e^(−2)=4*e^(−2)

Final Answer

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

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