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F(x)={sinx for x<22/7 {mx+b for x>22/7

  1. Feb 23, 2005 #1
    Hi any idea on this:
    Find all the values of the constants b and m for which the function.

    f(x)={sinx for x<22/7
    {mx+b for x>22/7

    I will appreciate for any help
     
  2. jcsd
  3. Feb 23, 2005 #2

    HallsofIvy

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    Rule 1: Copy the problem correctly! The way you've stated it, you can't find m and b. There exist functions having those values for all m and b! Did you mean "for which the function,..., IS DIFFERENTIABLE at x= 22/7"?

    In order to be differentiable at a point, the function must first be continuous there.
    Strictly speaking, the function you gave is not even defined at 22/7 and so can't be continuous there.

    I am going to assume that what you REALLY mean is: "Find all values of the constants b and m for which the function
    {sin x for x<= 22/7
    f(x)= { mx+ b for x> 22/7

    is differentiable." (Since f is clearly differentiable every where other than 22/7 the crucial part is that it be differentiable at 22/7).

    In order to be differentiable, it must first be continuous: the limit of f(x) at 22/7 FROM THE LEFT (x< 22/7) is sin(22/7) (And, once again: 22/7 is ALMOST pi. Did your problem really have 22/7 rather than pi). The limit of f(x) at 22/y FROM THE RIGHT is
    22m/7+ b so we must have 22m/7+ b= sin(22/7) (which is close to 0).

    The derivative of f, for x< 22/7 is cos(x). The derivative of f for x> 22/7 is m. While it is not necessary that derivatives be continuous, they do satisfy the "intermediate value problem" so if f is differentiable at 22/7, and since the derivative functions to the left and right of 22/7 are continuous, they must "match up" there: m= cos(22/7) (close to -1). You can put that into the other equation to solve for b. (In fact, mx+ b is simply the tangent line to sin(x) at x= 22/7.)
     
  4. Feb 23, 2005 #3
    sorry,sorry,sorry is true the one sent to me copied the problem wrong is like this

    Find all the values of the constants b and m for which the function.

    f(x)={sin x for x<pi
    {mx+b for x>=pi
    is a)continuous at x=pi b)is differentiable at x=pi
     
  5. Feb 24, 2005 #4

    HallsofIvy

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    Okay- in order fror the function to be continuous at pi, we must have
    [tex]m\pi+ b= sin(\pi)= 0[/tex]
    In order for the function to be differentiable at pi, we must also have
    [tex]m= cos(\pi)= -1[/tex].
     
  6. Feb 24, 2005 #5
    Thanx I understand u
     
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