## Carissa15 one year ago If anyone could me, I don't know what the mean value theorem is but I have a question about it as below.

1. Carissa15

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

2. freckles

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

3. Carissa15

I am not sure how to use the mean value theorem

4. freckles

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

5. freckles

and you are definitely going make use that c is in the interval (0,x) that is 0<c<x <--this will be handy inequality in your proof

6. freckles

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

7. freckles

anyways let me know if you still need help

8. 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?

9. freckles

no you have to show what it asked

10. freckles

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

11. freckles

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

12. freckles

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

13. 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

14. DanJS

It is an existence type thing

15. DanJS

|dw:1438761461750:dw|

16. DanJS

something like that..

17. Carissa15

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

18. Carissa15

Thank you all for your help

19. Carissa15

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

20. 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 0<c<x you will look at both parts 0<c and x>c.

21. Carissa15

Great, thank you so much. :-)