Find the absolute maximum and minimum values of a function f given by
`f(x)=2x^3-15 x^2+36 x+1` on the interval `[1, 5]`.

Explanation

`f(x)=2x^3−15x^2+36x+1`
diff on both side
`f ′(x)=6x^2−30x+36`
Putting `f ′ (x)=0`
`6x^2−30x+36=0`
`x^2−5x+6=0`
`(x−2)(x−3)=0`
`x=2 or 3`
We are given interval `[1,5]`
Hence, calculating f(x) at `2, 3, 1, 5,`
`f(2)=29`
`f(3)=28`
`f(1)=24`
`f(5)=56`
Hence, absolute maximum value is 56 at `x=5.`
Absolute minimum value is 24 at x=1.