Consider the function below. f(x)=x2e−x (a) Find the exact value of the minimum of f for x≥0. Find the exact value of the maximum of f for x≥0. (b) Find the exact value of x at which f increases most rapidly.

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Consider the function below.   f(x)=x2e−x  (a) Find the exact value of the minimum of f for x≥0.  Find the exact value of the maximum of f for x≥0.   (b) Find the exact value of x at which f increases most rapidly.

Joseph Fair · Accepted Answer

Step 1Given the functionf(x)=x2e−xStep 2First find the critical points, the first derivative of the function to zerof(x)=x2e−xf′(x)=e−xddxx2+x2ddxe−xf′(x)=2xe−x−x2e−xf′(x)=xe−x(2−x)f′(x)=0therefore,xe−x(2−x)=0x=0,x=2Step 3To find the exact maximum or minimum, plug the value of critical point into the function f(0)=0 f(2)=(2)2e−2=0.5413therefore, the exact value of the minimum f(x)=0  the exact value of the maximum f(x)=4e2

Karen Robbins · Accepted Answer

Step 4(b)To find the exact value of x at which f increases most rapidly.f′(x)=e−x(2x−x2)f″(x)=e−xddx(2x−x2)+(2x+x2)ddx(e−x)f″(x)=e−x(2−2x)+e−x(2x−x2)f″(x)=e−x(2−2x−2x+x2)f″(x)=e−x(x2−4x+2)f″(x)=0(x2−4x+2)=0x=2+2,x=2−2Plug these values into the first derivative and check whether the first derivative is positive and negativef′(2−2)=e−(2−2)[2(2−2)−(2−2)2] f′(2−2)=0.461 f′(2−2)&amp;gt;0 f′(2+2)=e−(2+2)[2(2+2)−(2+2)2]  f′(2+2)=−0.16  f′(2+2)=&amp;lt;0therefore, the function is increasing the most rapidly at x=(2−2)

Consider the function below. f(x)=x^{2}e^{-x} (a) Find the exact value of

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