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

1 Answer
Jul 8, 2018

Answer:

#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#