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Question

Find the points at which the function f given by f(x)=(x2)4(x+1)3 has local maxima

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Solution

f(x)=(x2)4(x+1)3

Derivative of given function is

f(x)=4(x2)3(x+1)3+3(x+1)2(x2)4

=(x+1)2(x3)3(7x2)

Putting this to zero we get x=1,2,27

duble derivative of this function is given by

f′′(x)=12(x2)2(x+1)3+12(x2)3(x+1)2+6(x+1)(x2)4+12(x+1)2(x2)3

At x=1 and 2 double derivative is 0 and hence function will not have any maxima or minima at these points.

At x=27

f′′(27)=12(272)2(27+1)3+12(272)3(27+1)2+6(27+1)(272)4+12(27+1)2(272)3

f′′(27)<0

So function will have local maxima at x=27

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