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Question

Find the local minimum and maximum values of f(x)=x43x3+3x2x.

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Solution

Given, f(x)=x43x3+3x2x
Thus, f(x)=4x39x2+6x1
At a turning point, f(x)=0 gives
4x39x2+6x1=0
(x1)2(4x1)=0x=1,1,14
When x=1,f(1)=0 and when x=14
f(14)=27256
Hence, the coordinates of the stationary points are
(1,0) and (14,27256)
f"(x)=12x218x+6
=6(2x23x+1)
=6(x1)(2x1)
x=1,f"(1)=0
Thus the second derivative test gives no information about the extremum nature of f at x=1.
When x=14,f"(14)=94>0.
Hence (14,27256) is a minimum point.

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