Question

Find the number c that satisfies the conclusion of the Mean Value Theorem. f(x)=x/x+2 Find the number c that satisfies the conclusion of the Mean Value Theorem. f(x)=x/x+2 @Mathematics

Answer 1

guess there is something missing here. ...

Answer 2

oh right sorry.
the interval is [1,4]

Answer 3

then you can do it. find
\[f(4),f(1),\frac{f(4)-f(1)}{3}\] and also find
\[f'(x)\] then set
\[f'(x)=\frac{f(4)-f(1)}{3}\] and solve for x

Answer 4

need help with the steps?

Answer 5

\[f(4)=\frac{4}{4+2}=\frac{2}{3}\]
\[f(1)=\frac{1}{1+2}=\frac{1}{3}\]
\[f(4)-f(1)=\frac{1}{3}\]
\[\frac{f(4)-f(1)}{3}=\frac{1}{9}\] check my arithmetic because i am typing it not write it

Answer 6

also you need
\[f'(x)=\frac{2}{(x+2)^2}\] and so now you set
\[\frac{2}{(x+2)^2}=\frac{1}{9}\] and solve for x

Answer 7

that's actually where my main problem is. i don't know how to solve for x at this point

Answer 8

ooooooooooooooooh

Answer 9

You're looking for the value of x where the slope of your function is equal to the slope of the dotted line. :)

Answer 10

solve as a ratio
\[\frac{2}{(x+2)^2}=\frac{1}{9}\]
\[(x+2)^2=18\]

Answer 11

i know it should be 18=(x+2)^2

Answer 12

yeah exactly

Answer 13

then done in two simple steps
\[x+2=\pm\sqrt{18}\]
\[x=-2\pm3\sqrt{2}\] but only one is in your interval

Answer 14

sometimes all it takes is asking and you see what you need to do before you get the answer

Answer 15

by the way these questions stink, because mvt says nothing about "find the c" it just says there is one

Answer 16

okay. thanks so much!! yeah they really do. do you think you could help me with 1 other one?

Answer 17

post sure why not

Answer 18

If f(3) = 7 and f '(x) ≥ 3 for 3 ≤ x ≤ 7, how small can f(7) possibly be?

Answer 19

ok this question looks weird but you can actually do it in your head

Answer 20

the very smallest f' can be is 3. that is like saying that you must go at least 3 units in the y for every unit in the x.

Answer 21

so if f(3) = 7, then what is the least f(7) can be? well 7 is 4 units from 3, so you must have increased at least 3*4 = 12 units, and 12 more than 7 is 19

the reason this is confusing is that the moron who wrote this question used 3 and 7 repeatedly

Answer 22

now we use mvt and write it out explicitly

Answer 23

we know there is a number c such that
\[\frac{f(7)-f(3)}{7-3}=f'(c)\geq 3\]
\[\frac{f(7)-7}{4}\geq 3\]
\[f(7)-7\geq 12\]
\[f(7)\geq 19\]

Answer 24

the way you said to to do it in your head makes perfect sense. thank you!!

Answer 25

it would have been much easier to understand if it said
\[f(2) =5,f'(x)\geq10\] and then asked what is the least
\[f(8)\] can be

Answer 26

you would say "from 2 to 8 is 6 units, and so we must have increased at least 6*10 = 60, and 60 more than 5 is 65"

Answer 27

i completely understand that. again, thanks so much.

Answer 28

yw