experimentX
  • experimentX
Math.
MIT 18.01 Single Variable Calculus (OCW)
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
experimentX
  • experimentX
it is given to prove that x/(1+x) < ln(1+x) < x in the interval [1,1+t] t>0
experimentX
  • experimentX
|dw:1350718620013:dw|
experimentX
  • experimentX
|dw:1350718731005:dw|

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experimentX
  • experimentX
|dw:1350718795042:dw|
Callisto
  • Callisto
But then, you still haven't considered lnx...
experimentX
  • experimentX
why do you need to prove ... ln(2) > 1/2 ?
Callisto
  • Callisto
It's not prove.. but.. we have to consider lnx..
experimentX
  • experimentX
why??
experimentX
  • experimentX
By considering ln x in [1; 1 + t] (t > 0), ln x in [1; 1 + t] (t > 0), minimum bound : ln(x) = 1 => x = e
experimentX
  • experimentX
x/(1+x) < ln(1+x) < x ... perhaps we can use the same argument as we used above.
experimentX
  • experimentX
\[ (x+1)^2 > 1+x > 1 \text{ for all x > 0} \\ {1 \over (1+x)^2} < {1 \over 1 +x} < 1 \text{ for all x > 0} \\ \int_0^x {1 \over (1+x)^2} \;dx < \int_0^x {1 \over 1 +x}\;dx < \int_0^x1 \; dx \\ {x \over 1+x} < \ln(1+x) < x \]
experimentX
  • experimentX
\[ \ln (x) \text{ in } [1, 1+t], t>0 \] perhaps this would mean show that \( \ln(1)<{t \over 1 +t} < \ln(1+t)\)
experimentX
  • experimentX
MVT http://upload.wikimedia.org/wikipedia/commons/thumb/e/ee/Mvt2.svg/300px-Mvt2.svg.png
experimentX
  • experimentX
|dw:1350722628100:dw|
experimentX
  • experimentX
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experimentX
  • experimentX
|dw:1350722732492:dw|
experimentX
  • experimentX
|dw:1350722797470:dw|
experimentX
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|dw:1350722973766:dw||dw:1350723003576:dw|
experimentX
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Callisto
  • Callisto
f(b) - f(a) / (b-a) ?
experimentX
  • experimentX
yep!!
experimentX
  • experimentX
|dw:1350724369177:dw|
Callisto
  • Callisto
OMG! My super smart brother!!
experimentX
  • experimentX
lol ... not that smart. Just copied half from your prev answer.

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