Find the maximum value of `2x^3-24 x+107` in the interval [1, 3]. Find the maximum value of the same function in `[3, 1]dot`

Explanation

Let `f(x)= 2x^3-24x+107`
`f'(x)=6x^2-24=6(x^2-4)`
Now, `f'(x)=0 implies 6(x^2-4)=0 implies x^2=4 implies x=pm2`
We first consider the interval` [1,3]` Then we evaluate the value of f at the critical point `x=2 `.
`f(2)= 2(8)-24(2)+107=16-48+107=75`
`f(1)=2(1)-24(1)+107=2-24+107=85`
`f(3)=2(27)-24(3)+107=54-72+107=89`
Hence the absolute maximum value of f(x) in the interval [1,3] is 89 occurring at `x=3`
Next , we consider the interval [-3,-1].
`f(-3)=2(-27)-24(-3)+107=-54+72+107=125`
`f(-1)=2(-1)-24(-1)+107=-2+24+107=129`
`f(-2)=2(-8)-24(-2)+107=-16+48+107=139`
Hence the absolute maximum value of f(x) in the interval `[-3,-1] `is 139 occurring at `x=2`