Question

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

Answer 1

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

Explanation:

Given -

`y=x^2-8x+12`

`dy/dx=2x-8`

`dy/dx=0 => 2x-8=0`

`x=8/2=4`

`(d^2y)/(dx^2)=2>0`

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

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