Question

What are the critical values of `f(x)=(3-x)^2-x`?

Answer 1

`x = 7/2`

Explanation:

Given: `f(x) = (3-x)^2 - x`

You find critical values by setting `f'(x) = 0` and solving for `x`.

Find the first derivative by using the power rule : `(u^n)' = n u^(n-1)u'`

Let `u = 3-x; " "u' = -1; " " n = 2`

`f'(x) = 2(3-x)(-1) - 1`

`f'(x) = -(6-2x) - 1`

`f'(x) = -6 + 2x - 1 = 2x - 7`

`f'(x) = 2x - 7 = 0`

Critical value: `" "2x = 7; " "x = 7/2`

Find the first derivative by using distribution :

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

`f(x) = 9 - 6x + x^2 - x`

`f(x) = x^2 -7x + 9`

`f'(x) = 2x - 7 = 0`

Critical value: `" "2x = 7; " "x = 7/2`