# Find the critical values

F(x) = x^5/7(x-5)^2

so you have to find the derivative and you end up with

f'(x) = x^5/7(2(x-5)) + 5/7x^-2/7(x-5)^2

simplified it to f'(x) = (19x^2-120x+125)/7x^2/7

so you use quadratic to find the zeroes of the numerator

which is 120 +/- sqrt(120^2-4(19)(125))/2(19)

simplifies out to (120 +/- 70)/38

so the zeros are 95/19 and 35/19

ive worked out the problem a couple times and came up with the same wrong answer.

## Answers and Replies

Mark44
Mentor
F(x) = x^5/7(x-5)^2
Your function is written correctly, but would be better written as (1/7)x^5/(x - 5)^2. Also, you started with F as the name of the function, so should keep the same letter to denote the derivative (i.e., not f').
so you have to find the derivative and you end up with
f'(x) = x^5/7(2(x-5)) + 5/7x^-2/7(x-5)^2
Mistake in the line above. It should be
F(x) = x^5/7(2(x-5)) + 5/7x^4/7(x-5)^2
simplified it to f'(x) = (19x^2-120x+125)/7x^2/7

so you use quadratic to find the zeroes of the numerator

which is 120 +/- sqrt(120^2-4(19)(125))/2(19)

simplifies out to (120 +/- 70)/38

so the zeros are 95/19 and 35/19

ive worked out the problem a couple times and came up with the same wrong answer.