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Slope of funtions graph

  1. Jun 1, 2014 #1
    I have to find the slope of the function

    g(x) = x/(x-2), (3,3)

    my attempt

    [(3+h)/((3+h)-2)] - [(3)/(3-2)] [itex]\div[/itex] h

    got rid of (3+h) and 3
    [(1/-2) -(1/-2)] [itex]\div[/itex] h
    0/h

    answer in book is -2
     
    Last edited: Jun 1, 2014
  2. jcsd
  3. Jun 1, 2014 #2

    adjacent

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    ##g(x)=\frac{x}{x-2}## right?
    Can't you use the quotient rule?
     
  4. Jun 1, 2014 #3
    yes g(x)= x/(x-2)

    I don't think I can use the quotient rule. Don't you need a limit? I wasn't given a limit only a function.
     
  5. Jun 1, 2014 #4

    adjacent

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    No.
    You can solve it in two ways.
    1. The definition of the derivative of a function
    2. Quotient rule
    The definition of the derivative of a function ##f## with respect to x: ##\frac{\text{d}f}{\text{d}x}=\lim_{\Delta x \to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}##

    The quotient rule states that ##(\frac{u}{v})'=\frac{u'v-uv'}{v^2}## Where u is the numerator and v is the denominator.

    The quotient rule is much easier.
     
  6. Jun 1, 2014 #5

    verty

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    ##{x \over x-2} \not = {1 \over -2}##.

    You may want to review fractions and what manipulations are allowed.
     
  7. Jun 1, 2014 #6
    Ok then I have to use the first option. Not allowed to use the second option yet.

    this is the definition I have to use

    The definition of the derivative of a function ##f## with respect to x: ##\frac{\text{d}f}{\text{d}x}=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}##
     
  8. Jun 1, 2014 #7

    adjacent

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    Your attempt:$$\frac{\frac{3+h}{(3+h)-2}-\frac{3}{3-2}}{h}$$ is correct.
    You only have some algebra problems in simplifying that. Try again.

    If you use that definition,you will get the derivative with respect to x. If you use 3 instead of x, you will get derivative of the function at x=3. That's what you did there in your attempt
     
    Last edited: Jun 1, 2014
  9. Jun 1, 2014 #8
    thank all of you who helped I have figured it out. I do not need help anymore.
     
  10. Jun 1, 2014 #9

    adjacent

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    Happy to know that :smile:
     
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