Question

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!

Answer 1

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

Answer 2

yes

Answer 3

which x value makes the denominator zero?

Answer 4

-6

Answer 5

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

Answer 6

no

Answer 7

so the function is continuous on [1,12]

that allows us to use the MVT properly here

Answer 8

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

Answer 9

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

Answer 10

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

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

Answer 11

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

Answer 12

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

Answer 13

you said `that gives me 1?`

what do you mean exactly? that's the secant slope you got?

Answer 14

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

Answer 15

that's not the correct secant slope

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

yes

Answer 20

okay I got that part!

Answer 21

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

Answer 22

now you need f ' (x)

Answer 23

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

Answer 24

no

Answer 25

you set them equal to one another and solve for x

Answer 26

I set what equal to each other?

Answer 27

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

Answer 28

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

Answer 29

what would i do w the other x?

Answer 30

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

Answer 31

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

Answer 32

those are the correct solutions to the equation

you now need to check if they lie in the interval [1,12]

Answer 33

okay so it would only be -6+3root14

Answer 34

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

Answer 35

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

Answer 36

thank you so much!!