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Question

Let f:[1,1]R, where f(x)=2x3x410. The minimum value of f(x) is
  1. -13

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Solution

The correct option is A -13
f(x)=2x3x410,
f(x)=6x24x3,
f′′(x)=12x12x2,
f′′′(x)=1224x
For CriticalPoints;f(x)=0
6x24x3=0
x=0,x=32 are stationary points
x=32/ϵ[1,1] So we have only one point i.e.,x=0
f′′(0)=0 and f′′(0)0
So x=0 is a point of inflection
So Minimum value occur only at corner points
i.e.,f(1)=13&f(1)=9
So Min. f(x)=13.

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