Find the absolute maximum and absolute minimum vlaues of a function f given by `f(x)=2x^(3)-15x^(2)+36x+3` on the interval [0,6] .

Explanation

We have `f(x)=2x^(3)-15x^(2)+36x+3`
`f'(x)=6x^(2)-30x+36`
`=6(x-3)(x-2)`
For maxima and minima
`f'(x)=0`
(x-2)(x-3)=0
x=2,3
Now using rule to find absolute maxima and absolute minima
f(0)=3
`f(x)=2(2)^(3)-15(2)^(2)+36(2)+3=31`
`f(3)=2(3)^(3)-15(3)^(2)+36(3)+3=30`
`f(6)=2(6)^(3)-15(6)^(2)+36xx6+3=111`
Hence Absolute maxima of function =f(6)=111
Absolute minima of function =f(0)=3