Question

What are the absolute extrema of `f(x)=(x-2)(x-5)^3 + 12in[1,4]`?

Answer 1

Local Minima. is `-2187/128.`
Global Minima`=-2187/128~=-17.09.`
Global Maxima `=64.`

Explanation:

For extrema, `f'(x)=0.`

`f'(x)=(x-2)*3(x-5)^2+(x-5)^3*1=(x-5)^2{3x-6+x-5]=(4x-11)(x-5)^2.`

`f'(x)=0 rArr x=5!in [1,4],` so no need for further cosideration & `x=11/4.`

`f'(x)=(4x-11)(x-5)^2, rArr f''(x)=(4x-11)*2(x-5)+(x-5)^2*4=2(x-5){4x-11+2x-10}=2(x-5)(6x-21).`

Now, `f''(11/4)=2(11/4-5)(33/2-21)=2(-9/4)(-9/2)>0,` showing that, `f(11/4)=(11/4-2)(11/4-5)^3=(3/2)(-9/4)^3=-2187/128,` is Local Minima.

To find Global Values, we need `f(1)=(1-2)(1-5)^3=64,` & `f(4)=(4-2)(4-5)^3=-2.`

Hence, Global Minima `=min` { local minima, `f(1), f(4)}=min{-2187/128,64, -2}=min{-17.09, 64, -2}=-2187/128~=-17.09`

Global Maxima `=max` { local maxima ( which doesn't exist), `f(1), f(4)}=max{64, -2}=64.`