Find the critical values

  • Thread starter apiwowar
  • Start date
  • #1
96
0
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.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
35,062
6,796
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.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Related Threads on Find the critical values

  • Last Post
Replies
2
Views
7K
  • Last Post
Replies
5
Views
1K
Replies
1
Views
5K
Replies
1
Views
881
  • Last Post
Replies
4
Views
975
Replies
1
Views
471
Replies
0
Views
917
Replies
4
Views
3K
Replies
1
Views
4K
Top