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

Apr 15, 2018

There's no global maxima.

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

#### Explanation:

$f \left(x\right) = \frac{{x}^{3} - 7 {x}^{2} + 12 x - 6}{x - 1}$

$f \left(x\right) = \frac{\left(x - 1\right) \left({x}^{2} - 6 x + 6\right)}{x - 1}$

$f \left(x\right) = {x}^{2} - 6 x + 6 ,$where x≠ 1

$f ' \left(x\right) = 2 x - 6$

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

Endpoints: 1 & 4:

$x = 1$
$f \left(1\right) : \text{undefined}$
lim_(x→1) f(x) = 1

$x = 4$
$f \left(4\right) = - 2$

Critical point(s):
$f ' \left(x\right) = 2 x - 6$
$f ' \left(x\right) = 0$
$2 x - 6 = 0 , x = 3$

At $x = 3$
$f \left(3\right) = - 3$

There's no global maxima.

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