anonymous
  • anonymous
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!
Mathematics
schrodinger
  • 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
  • jim_thompson5910
the function is \[\Large f(x) = \frac{x}{x+6}\] right?
anonymous
  • anonymous
yes
jim_thompson5910
  • 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
  • anonymous
-6
jim_thompson5910
  • jim_thompson5910
is -6 in the interval [1,12] ?
anonymous
  • anonymous
no
jim_thompson5910
  • jim_thompson5910
so the function is continuous on [1,12] that allows us to use the MVT properly here
jim_thompson5910
  • jim_thompson5910
if there was a discontinuity on [1,12] then we couldn't use the MVT
anonymous
  • 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
  • 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
  • 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
  • anonymous
so i would substitute 1 for a and 12 for b? that gives me 1?
jim_thompson5910
  • jim_thompson5910
you said `that gives me 1?` what do you mean exactly? that's the secant slope you got?
anonymous
  • 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
  • jim_thompson5910
that's not the correct secant slope
jim_thompson5910
  • jim_thompson5910
\[\Large f(x) = \frac{x}{x+6}\] \[\Large f(1) = \frac{1}{1+6}\] \[\Large f(1) = \frac{1}{7}\]
jim_thompson5910
  • 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
  • anonymous
and then i would do the same for f(12)?
jim_thompson5910
  • jim_thompson5910
yes
anonymous
  • anonymous
okay I got that part!
jim_thompson5910
  • 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
  • jim_thompson5910
now you need f ' (x)
anonymous
  • anonymous
so then i would substitute 1/21 for x in this 6/(6+x)^2 right?
jim_thompson5910
  • jim_thompson5910
no
jim_thompson5910
  • jim_thompson5910
you set them equal to one another and solve for x
anonymous
  • anonymous
I set what equal to each other?
jim_thompson5910
  • jim_thompson5910
1/21 and that f ' (x) you got
anonymous
  • anonymous
i end up with 6=1/21x^2+4/7x+12/7
anonymous
  • anonymous
what would i do w the other x?
jim_thompson5910
  • 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
  • anonymous
got -6+3root14 and -6-3root 14?
jim_thompson5910
  • jim_thompson5910
those are the correct solutions to the equation you now need to check if they lie in the interval [1,12]
anonymous
  • anonymous
okay so it would only be -6+3root14
jim_thompson5910
  • jim_thompson5910
yes since \(\large -6+3\sqrt{14} \approx 5.22\) which is in the interval [1,12]
jim_thompson5910
  • jim_thompson5910
the other value is -17.22 which is not in that interval
anonymous
  • anonymous
thank you so much!!

Looking for something else?

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