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`
` rArr f'(x) = 6x^(2)-24 = 6(x^(2)-4)=6(x+2)(x-2)`
For maxima/minima, f'(x) = 0
` rArr 6(x+2)(x-2)=0`
` rArr x=2, -2, `
We find the values of `f(x) at x = 2 in [1, 3]` and at the end points of the interval [1, 3].
at ` x= 1, f(1) = 2 xx 1^(3) - 24 xx 1 + 107 = 85 `
at ` x = 2, f(2) = 2 xx 2^(3) - 24 xx 2 + 107 = 75`
at ` x = 3, f(3) = 2 xx 3^(3) - 24 xx 3 + 107 = 89`
So the absolute maximum value of f(x) in the interval [1, 3] is 89, which is at x = 3.
Now, we find the values of `f(x) at x=-2 in [-3, -1] ` and at the end points of the interval [-3, -1].
at ` x =- 1, f(-1)= 2 (-1)^(3) - 24 (-1) + 107`
` =- 2 + 24 +107 = 129`
at ` x = - 2, f(-2) = 2 (-2)^(3) - 24 (-2)+107`
` =- 16 + 48+107 = 139`
at ` x =- 3, f(-3), f(-3)= 2 (-3)^(3) - 24 (-3)+107`
` =- 54 + 72 + 107 = 125`
So the absolute maximum value of f(x) in the interval [-3, -11] is 139 which is at x = - 2.