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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.

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  1. Carissa15
    • one year ago
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    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
    • one year ago
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    \[\text{ try something like this } \\ \text{ use the mvt for } f(x)=\ln(1+x) \text{ on } (0,x)\]

  3. Carissa15
    • one year ago
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    I am not sure how to use the mean value theorem

  4. freckles
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    the function f(x)=ln(x+1) is continuous on [0,x] and differentiable on (0,x)*

  7. freckles
    • one year ago
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    anyways let me know if you still need help

  8. Carissa15
    • one year ago
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    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
    • one year ago
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    no you have to show what it asked

  10. freckles
    • one year ago
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    use the thing above find f'(c) and...f(x) and f(0)

  11. freckles
    • one year ago
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    \[f(x)=\ln(x+1)\\ \text{ can you find } f'(x)?\]

  12. freckles
    • one year ago
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    not sure if you are there or not but I have to go

  13. DanJS
    • one year ago
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    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
    • one year ago
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    It is an existence type thing

  15. DanJS
    • one year ago
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    |dw:1438761461750:dw|

  16. DanJS
    • one year ago
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    something like that..

  17. Carissa15
    • one year ago
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    thank you, i will have to come back to this. thanks

  18. Carissa15
    • one year ago
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    Thank you all for your help

  19. Carissa15
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    Great, thank you so much. :-)

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