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(MVT) f(x)=sinx. Show that, for any given a and b, |sina-sinb|<=|b-a|

  1. Nov 15, 2011 #1
    I'm so bad with the Mean Value Theorem. Can someone help me prove that, if f(x)=sinx, that, for any given a and b, |sinb-sina|<=|b-a|. Explain if you could too, please. Thanks a lot.
     
  2. jcsd
  3. Nov 15, 2011 #2

    gb7nash

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    Start with what you know. What does the mean value theorem say?
     
  4. Nov 15, 2011 #3
    First of all, I just realized I should've posted this in the homework section. And I know the function is cont. on [a,b] and f'(x) exists on (a,b) so f'(c)=(f(b)-f(a))/(b-a). So f'(c)=(sinb-sina)/(b-a). But I'm not sure how f'(c) comes in to play. And I don't know how to interperate the absolute value either.
     
  5. Nov 15, 2011 #4

    gb7nash

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    Ok, so we have:

    (sin(b)-sin(a)) = f'(c)(b-a)

    It's a well-known fact that x = y → |x| = |y|.

    How can we apply that?
     
  6. Nov 15, 2011 #5
    I honestly don't know...lol :\
     
  7. Nov 15, 2011 #6

    gb7nash

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    This says that:

    left-hand side = right-hand side

    can be rewritten as:

    |left-hand side| = |right-hand side|

    For example:

    a+b = c+d → |a+b| = |c+d|

    Now, let's look at the problem at hand. We have:

    (sin(b)-sin(a)) = f'(c)(b-a)

    Applying the absolute value on both sides, how can we rewrite this?
     
  8. Nov 15, 2011 #7
    |sin(b)-sin(a)| = |f'(c)(b-a)|. I understand this, I just read it wrong, lol sorry
     
  9. Nov 15, 2011 #8

    gb7nash

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    Ok, good. Now, let's calculate what f'(c) is. What's f'(x)? After you have that, plug c in.
     
  10. Nov 15, 2011 #9
    f'(x) = cosx so f'(c) = cos(c).
     
  11. Nov 15, 2011 #10

    gb7nash

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    Correct. So now, we have:

    |sin(b)-sin(a)| = |cos(c)(b-a)|

    It's also a well-known fact that |xy| = |x||y|. How could we rewrite the right-hand side?
     
  12. Nov 15, 2011 #11
    |sin(b)-sin(a)| = |cos(c)||b-a|. And -1<=cosx<=1 so would it be |sin(b)-sin(a)| <= 1|b-a| ?
     
  13. Nov 15, 2011 #12

    gb7nash

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    Correct.
     
  14. Nov 15, 2011 #13
    Thank you so much for working with me and not just posting an unexplained proof. Props!
     
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