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!

- anonymous

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- jim_thompson5910

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

- anonymous

yes

- jim_thompson5910

which x value makes the denominator zero?

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## More answers

- anonymous

-6

- jim_thompson5910

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

- anonymous

no

- jim_thompson5910

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

- jim_thompson5910

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

- anonymous

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

- 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
\[\Large f \ '(c) = \frac{f(b) - f(a)}{b-a}\]

- 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

- anonymous

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

- jim_thompson5910

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

- 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

- jim_thompson5910

that's not the correct secant slope

- jim_thompson5910

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

- jim_thompson5910

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

- anonymous

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

- jim_thompson5910

yes

- anonymous

okay I got that part!

- jim_thompson5910

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

- jim_thompson5910

now you need f ' (x)

- anonymous

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

- jim_thompson5910

no

- jim_thompson5910

you set them equal to one another and solve for x

- anonymous

I set what equal to each other?

- jim_thompson5910

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

- anonymous

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

- anonymous

what would i do w the other x?

- 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

- anonymous

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

- jim_thompson5910

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

- anonymous

okay so it would only be -6+3root14

- jim_thompson5910

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

- jim_thompson5910

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

- anonymous

thank you so much!!

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