Question

What are the absolute extrema of `f(x)=(x^3-7x^2+12x-6)/(x-1)in[1,4]`?

Answer 1

There's no global maxima.

The global minima is -3 and occurs at x = 3.

Explanation:

`f(x) = (x^3 - 7x^2 + 12x - 6)/(x - 1)`

`f(x) = ((x - 1)(x^2 - 6x + 6)) /(x - 1)`

`f(x) = x^2 - 6x + 6,`where `x≠ 1`

`f'(x) = 2x - 6`

The absolute extrema occurs on an endpoint or at the critical number.

Endpoints: `1 & 4:`

`x = 1`
`f(1) : "undefined"`
`lim_(x→1) f(x) = 1`

`x = 4`
`f(4) = -2`

Critical point(s):
`f'(x) = 2x - 6`
`f'(x) = 0`
`2x - 6 = 0 , x = 3`

At `x = 3`
`f(3) = -3`

There's no global maxima.

There's no global minima is -3 and occurs at x = 3.