## anonymous one year ago Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval? f(x) = x/x+6 [1, 12] If it satisfies the hypotheses, find all numbers c that satisfy the conclusion of the Mean Value Theorem. (Enter your answers as a comma-separated list. If it does not satisfy the hypotheses, enter DNE). Ive tried to do it multiple ways but i cant get the right answer. Please help!

1. jim_thompson5910

the function is $\Large f(x) = \frac{x}{x+6}$ right?

2. anonymous

yes

3. jim_thompson5910

which x value makes the denominator zero?

4. anonymous

-6

5. jim_thompson5910

is -6 in the interval [1,12] ?

6. anonymous

no

7. jim_thompson5910

so the function is continuous on [1,12] that allows us to use the MVT properly here

8. jim_thompson5910

if there was a discontinuity on [1,12] then we couldn't use the MVT

9. anonymous

okay, I understand that part, the equation part is whats given me a lot of trouble, i dont have examples like this one

10. jim_thompson5910

Mean Value Theorem If f(x) is continuous on [a,b] and differentiable on (a,b), then there exists at least one value of c such that $\Large f \ '(c) = \frac{f(b) - f(a)}{b-a}$

11. jim_thompson5910

what you have to do is find the secant slope through (1, f(1)) and (12,f(12)) then find f ' (x) set f ' (x) equal to the secant slope and solve for x

12. anonymous

so i would substitute 1 for a and 12 for b? that gives me 1?

13. jim_thompson5910

you said that gives me 1? what do you mean exactly? that's the secant slope you got?

14. anonymous

yeah if i substitute the a for 1 and 12 for b, that equals to 11/11 so 1? i dont think im doing this right

15. jim_thompson5910

that's not the correct secant slope

16. jim_thompson5910

$\Large f(x) = \frac{x}{x+6}$ $\Large f(1) = \frac{1}{1+6}$ $\Large f(1) = \frac{1}{7}$

17. jim_thompson5910

$\Large f(x) = \frac{x}{x+6}$ $\Large f(12) = \frac{12}{12+6}$ $\Large f(12) = \frac{12}{18}$ $\Large f(12) = \frac{2}{3}$

18. anonymous

and then i would do the same for f(12)?

19. jim_thompson5910

yes

20. anonymous

okay I got that part!

21. jim_thompson5910

you should have this next $\Large f \ '(c) = \frac{f(b) - f(a)}{b-a}$ $\Large f \ '(c) = \frac{f(12) - f(1)}{12-1}$ $\Large f \ '(c) = \frac{\frac{2}{3} - \frac{1}{7}}{12-1}$ $\Large f \ '(c) = \frac{\frac{11}{21}}{11}$ $\Large f \ '(c) = \frac{1}{21}$

22. jim_thompson5910

now you need f ' (x)

23. anonymous

so then i would substitute 1/21 for x in this 6/(6+x)^2 right?

24. jim_thompson5910

no

25. jim_thompson5910

you set them equal to one another and solve for x

26. anonymous

I set what equal to each other?

27. jim_thompson5910

1/21 and that f ' (x) you got

28. anonymous

i end up with 6=1/21x^2+4/7x+12/7

29. anonymous

what would i do w the other x?

30. jim_thompson5910

$\Large f \ '(x) = \frac{6}{(x+6)^2}$ $\Large \frac{1}{21} = \frac{6}{(x+6)^2}$ $\Large 1*(x+6)^2 = 21*6$ $\Large (x+6)^2 = 126$ keep going to solve for x

31. anonymous

got -6+3root14 and -6-3root 14?

32. jim_thompson5910

those are the correct solutions to the equation you now need to check if they lie in the interval [1,12]

33. anonymous

okay so it would only be -6+3root14

34. jim_thompson5910

yes since $$\large -6+3\sqrt{14} \approx 5.22$$ which is in the interval [1,12]

35. jim_thompson5910

the other value is -17.22 which is not in that interval

36. anonymous

thank you so much!!