If anyone could me, I don't know what the mean value theorem is but I have a question about it as below.

- Carissa15

- katieb

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- Carissa15

Apply the mean value Theorem to f(x)=In(1+x) to show that \[\frac{ x }{ 1+x }<\ln (1+x)0.\]

- freckles

\[\text{ try something like this } \\ \text{ use the mvt for } f(x)=\ln(1+x) \text{ on } (0,x)\]

- Carissa15

I am not sure how to use the mean value theorem

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## More answers

- freckles

the function f(x)=ln(x+1) is continuous on [0,1] and differentiable on (0,1)
so there exist
\[c \in (0,x) \text{ such that } f'(c)=\frac{f(x)-f(0)}{x-0}\]

- freckles

and you are definitely going make use that c is in the interval (0,x)
that is 0

- freckles

the function f(x)=ln(x+1) is continuous on [0,x] and differentiable on (0,x)*

- freckles

anyways let me know if you still need help

- Carissa15

Thank you, so all it is asking is to prove that the function will be continuous and differentiable from the values using the mvt?

- freckles

no you have to show what it asked

- freckles

use the thing above
find f'(c)
and...f(x) and f(0)

- freckles

\[f(x)=\ln(x+1)\\ \text{ can you find } f'(x)?\]

- freckles

not sure if you are there or not
but I have to go

- DanJS

The mean value theorem basically says, if you have a continuous function between two x values,
There is some point in between the interval ends where the slope of the tangent line to the function is the same as the slope of a secant line connecting the endpoints...if i remember right

- DanJS

It is an existence type thing

- DanJS

|dw:1438761461750:dw|

- DanJS

something like that..

- Carissa15

thank you, i will have to come back to this. thanks

- Carissa15

Thank you all for your help

- Carissa15

I am still unsure of this, I think I need to substitute any x>0 for each of the 3 equations \[\frac{ x }{ 1+x }<\ln(1+x) 0 each equation will be less than the last?

- freckles

we can use that formula in the mvt since our function satisfies that
f is continuous on [0,x] and differentiable on (0,x)
I posted above this means we have some number c in (0,x) such that:
\[f'(c)=\frac{f(x)-f(0)}{x-0} \\ \text{ where } f(x)=\ln(x+1)\]
You need to find f'(x)...
Then plug in c into that so you can have your left hand side.
Then evaluate f(0) and replace f(0) with whatever you get for that.
Then use the inequality 0c.

- Carissa15

Great, thank you so much. :-)

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