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Prove f(x) is differentiable at x=1 where f(x)=2x^2 x=<1, =4x-1 x>1

  1. Nov 11, 2007 #1
    prove f(x) is differentiable at x=1:
    f(x)=2x^2 x(less than or equal to)1
    4x-1 x>1
     
  2. jcsd
  3. Nov 11, 2007 #2

    PhY

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    do you mean:
    d/dx:= d/dx(2x^2x) * d/dx(2x) ala function of a function?

    i'm probably wrong.
     
  4. Nov 11, 2007 #3
    no, i mean it's in piecewise form with the big curly brackets
     
  5. Nov 11, 2007 #4

    PhY

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    You Mean Integration.

    Argh, My Integration is Rusty.
    its the opposite of Differentiation.

    so 2x would be x^2
    2x^2x ...i don't know, because its F(X)^G(X).

    You need to hear from somebody else on this.
     
  6. Nov 11, 2007 #5
    yeah i think i do need to hear from somebody else on this cos u didnt understand the question ;)
    i have to prove that it is differentiable at x=1 it has nothing to do with integration.
    thanks anyway
     
  7. Nov 11, 2007 #6

    cristo

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    The derivative of a function at a point can be expressed as the limit of an expression. You should be able to get two limits; one for each branch of the function. If these are the same, then the function is differentiable at x=1.
     
  8. Nov 11, 2007 #7

    JasonRox

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    It's amazing how complicated one can make a question when it is meant to be simple. (Not meant to the OP.)
     
  9. Nov 11, 2007 #8
    thanx v much i think i can do it now.
     
  10. Nov 11, 2007 #9

    CompuChip

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    You might also want to check that it is continuous. For example,
    [tex]f(x) = \left\{
    \begin{array}{rl}
    0 & \text{ if } x \le 0 \\
    1 & \text{ if } x > 0
    \end{array}
    [/tex]
    will give you 0 for the derivative when approaching from the left or right to zero, though at x = 0 the function is not continuous at all.
     
    Last edited: Nov 11, 2007
  11. Nov 11, 2007 #10

    HallsofIvy

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    No, he meant differentiation. Determine whether
    [tex]f(x) = \left\{ \begin{array}{rl} 2x^2 & \text{ if } x \le 1 \\ 4x-1 & \text{ if } x > 1 \end{array}[/tex]
    is differentiable at x= 1

    Compuchip's suggestion is etremely good here: a function can't be differentiable if it is't continuous at the point! What is the limit of f as you approach 1 from the left? What is the limit as you approach from the right?
     
  12. Nov 12, 2007 #11

    Gib Z

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    Or more precisely, [tex]\frac{d}{dx}\int^x_a f(t) dt = f(x)[/tex] where a is a constant. "The Opposite of differentiation" is what people told me before I started integral calculus as well, and it screwed up my understanding a heap load.

    Just in case you want to know, you let y=2x, and express the remaining integral in terms of the exponential and logarithmic functions.
     
  13. Nov 12, 2007 #12

    CompuChip

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    I still don't see why you would want to do integration.
    It's a partwise defined function, all you need to do is show that the function and the derivative are continuous.
     
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