Locate the critical points of the following function. Then use the Second Deriva

Question

Locate the critical points of the following function. Then use the Second Derivative Test to determine whether they correspond to local maxima, local minima, or neither. f(x) = 3x^2 e^-2x -1 What is(are) the critical point(s) of f? Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The critical point(s) is(are) x =. (Use a comma to separate answers as needed.) B. There are no critical points for f. What is/are the local minimum/minima of f? Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The local minimum/minima of f is/are at x = B. There is no local minimum of f. What is/are the local maximum/maxima of f? Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The local maximum/maxima of f is/are at x =. B. There is no local maximum of f.

Explanation / Answer

given f(x)=3x2e-2x-1

differentiate with respect to x

f '(x)=6xe-2x +(3x2)(-2)e-2x-0

f '(x)=6xe-2x -6x2e-2x

for critical point f '(x) =0

6xe-2x -6x2e-2x=0

6x-6x2=0

6x(1-x)=0

x=0,x=1

critical points are x =0,1

f ''(x)=6e-2x+6x(-2)e-2x -(12xe-2x+6x2(-2)e-2x)

f ''(x)=6e-2x-12xe-2x -12xe-2x+12x2e-2x

f ''(x)=(6-24x+12x2)e-2x

f "(0)=6>0 so local minimum at x =0

f ''(1)=(6-24+12)e-2=-6/e2<0

so local maximum at x =1

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