# What are the extrema of f(x)=x^2 - 8x + 12 on [-2,4]?

Aug 21, 2016

the function has a minimum at $x = 4$ graph{x^2-8x+12 [-10, 10, -5, 5]}

#### Explanation:

Given -

$y = {x}^{2} - 8 x + 12$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x - 8$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 0 \implies 2 x - 8 = 0$

$x = \frac{8}{2} = 4$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 2 > 0$

At x=4; dy/dx=0;(d^2y)/(dx^2)>0

Hence the function has a minimum at $x = 4$