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Exercise with lagrange and derivatives

  • Thread starter Felafel
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  • #1
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Homework Statement



Being a>0 and f:[a,b]--->R continuos and differentiable in (a,b), show that there exists a t ##\in## (a,b) such that:
## \frac{bf(a)-af(b)}{b-a}=f(t)-tf'(f)##

The Attempt at a Solution


For lagrange's theorem, we have:
## \frac{f(a)-f(b)}{b-a}= -f'(t) ##

thought i could find f(t) from the equation ##f(x) = f(t) + f'(t)(x-t)+R_1(x-t)## ignoring R.
but then a "x" would appear and I don't know how to deal with it.
 
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Answers and Replies

  • #2
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To simplify the equation, consider g:[a,b]->R with
$$g(x)=f(x)-f(a)-(x-a)\frac{f(b)-f(a)}{b-a}$$
This is chosen to get g(a)=g(b)=0 and ##g'(x)=f'(x)-\frac{f(b)-f(a)}{b-a}##

Is there some t ∈ (a,b) such that ##0=g(t)-tg'(t)##?
Working backwards, you should be able to get the original equation.
 
  • #3
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sounds good, but I'm explicitly asked to use Lagrange's theorem
 
  • #4
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I think you can use that theorem to solve the reduced problem ;).
 
  • #5
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another hint? i don't get it, i always end up in a loop :(
 

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