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Tangent line to the curve f(x)= x + 1/x

  1. Sep 17, 2011 #1
    1. The problem statement, all variables and given/known data
    So I have to find the tangent line when x=5. I use the difference quotient, but I'm stuck at the algebra? I can't seem to figure out what steps I should take.

    Here's what I have:

    [(x+h)+1/(x+h)] - (x+1/x)

    I've tried numerous steps on how to solve this, but can't do so. What I've basically have been trying is is to get the common denominators for the left and right sides, and then get another common denominator for both sides. What am I doing wrong?

    Thank you in advance.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 17, 2011 #2

    micromass

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    Did you see derivatives, yet?? Because that's what you need to use.
     
  4. Sep 17, 2011 #3

    gb7nash

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    Besides the limit, you're missing one thing. You want:

    [tex]\frac{f(x+h)-f(x)}{h}[/tex]

    So really, you have:

    [tex]\frac{(x+h)+\frac{1}{x+h} - (x+\frac{1}{x})}{h}[/tex]

    Simplify this a bit and on the *numerator part, find a common denominator like you were doing before.
     
  5. Sep 17, 2011 #4
    Yeah we're on derivatives, I just forgot to put the whole equation over h.

    @gb7nash That's where I'm stuck, when find a common denominator and subtract both equations, the h doesn't cancel out. So I'm doing something wrong.

    When I add the x+h and 1/x+h, I get [(x+h)^2 + 1]/(x+h) - (x^2+1)/x
    Now do I get another common denominator?
     
  6. Sep 17, 2011 #5

    gb7nash

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    Yes. Before you do anything with that, simplify the numerator. We can cancel something out.
     
  7. Sep 17, 2011 #6
    @gb7nash

    Okay, so now I have: [x^2+2xh+h^2+1]/(x+h) - (x^2+1)/x

    Get another com. denom then I have: [x(x^2+2xh+h^2+1) - (x^2+1)] / x(x+h)
     
  8. Sep 17, 2011 #7

    gb7nash

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    I'm not quite sure what you're trying to do.

    [tex]\frac{(x+h)+\frac{1}{x+h} - (x+\frac{1}{x})}{h} = \frac{h+\frac{1}{x+h} - \frac{1}{x}}{h}[/tex]

    Now, looking at the numerator, we want to find a common denominator for [itex]\frac{1}{x+h}[/itex] and [itex]\frac{1}{x}[/itex]

    How can we do this?
     
  9. Sep 17, 2011 #8
    Ohh, I was trying to add up all the fractions first. I didn't even notice the x's canceled out. Okay, so when I subtract 1/x+h - 1/x I get -h/x+h

    After the mess, I get my answer to be (h+x-1)/x+h
     
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