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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.
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.

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Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0Apply the mean value Theorem to f(x)=In(1+x) to show that \[\frac{ x }{ 1+x }<\ln (1+x)<x, for x>0.\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[\text{ try something like this } \\ \text{ use the mvt for } f(x)=\ln(1+x) \text{ on } (0,x)\]

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0I am not sure how to use the mean value theorem

freckles
 one year ago
Best ResponseYou've already chosen the best response.3the function f(x)=ln(x+1) is continuous on [0,1] and differentiable on (0,1) so there exist \[c \in (0,x) \text{ such that } f'(c)=\frac{f(x)f(0)}{x0}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3and 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

freckles
 one year ago
Best ResponseYou've already chosen the best response.3the function f(x)=ln(x+1) is continuous on [0,x] and differentiable on (0,x)*

freckles
 one year ago
Best ResponseYou've already chosen the best response.3anyways let me know if you still need help

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0Thank you, so all it is asking is to prove that the function will be continuous and differentiable from the values using the mvt?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3no you have to show what it asked

freckles
 one year ago
Best ResponseYou've already chosen the best response.3use the thing above find f'(c) and...f(x) and f(0)

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[f(x)=\ln(x+1)\\ \text{ can you find } f'(x)?\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3not sure if you are there or not but I have to go

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1The 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
 one year ago
Best ResponseYou've already chosen the best response.1It is an existence type thing

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0thank you, i will have to come back to this. thanks

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0Thank you all for your help

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0I 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?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3we 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: \[f'(c)=\frac{f(x)f(0)}{x0} \\ \text{ where } f(x)=\ln(x+1)\] 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.

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0Great, thank you so much. :)
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