- #1

- 96

- 0

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.