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

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- jim_thompson5910

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

- anonymous

yes

- jim_thompson5910

which x value makes the denominator zero?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## 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!!

Looking for something else?

Not the answer you are looking for? Search for more explanations.