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Simple question, tangent line

  1. Aug 22, 2006 #1


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    really isnt calculus but thats the name of my class so, anyway im having a problem here. i think its pretty simple im just missing something. im supposed to find the slope of the tangent line using one point and:

    slope of tangent line =



    find slope of tangent line:


    at a=2



    common denominator X by (1+h)






    make "h" zero and im stuck.....
  2. jcsd
  3. Aug 22, 2006 #2
    Ummm... Is 1/2 the same as (1+h)/(2+h)?
  4. Aug 22, 2006 #3


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    okay stupid mistake, now im stuck again tho


    comes out to


    gives me -1/h which doesnt work, darn
  5. Aug 22, 2006 #4
    You need a factor of h in the first denominator for the denominators to be equal.
  6. Aug 22, 2006 #5


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    i dont understand
  7. Aug 22, 2006 #6
    The denominator of the first term expanded is (4+2h) while the second term's denominator is (4h+2h2)
  8. Aug 22, 2006 #7


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    but those are different terms, arent i looking for a common denominator?
  9. Aug 22, 2006 #8


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    Homework Helper

    If [tex]f(x)=\frac{1}{x},[/tex] then

    [tex]\frac{f(a+h)-f(a)}{h} = \frac{\frac{1}{2+h}-\frac{1}{2}}{h} = \frac{1}{h(2+h)}-\frac{1}{2h} = \frac{2-(2+h)}{2h(2+h)} [/tex]
    [tex]=-\frac{1}{2(2+h)}\rightarrow -\frac{1}{4} \mbox{ as }h\rightarrow 0[/tex]
  10. Aug 22, 2006 #9
    Yes which is why you need to multiply the first term by h/h so that you have a common denominator.
  11. Aug 22, 2006 #10


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    thanks for the replies, but i dont understand how the c.d is 2h(2+h)
  12. Aug 24, 2006 #11


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    finally, got it
  13. Aug 24, 2006 #12
    I was wondering one thing about tangent line. Since tangent lines can also be secant line, I actually had a question about secant line. Is it possible for a secant line to intersect 2 consecutive points of a curve? I guess what I am trying to get to is, in a curve, can two, next to each other points be in a straight line?
  14. Aug 24, 2006 #13
    What do you mean by two consecutive points? Between any two points on a curve there are an infinite number of other points so there really isn't such a thing as consecutive points in the way you are probably thinking. But if you mean like 2 points on a curve where say x = 1 and x = 2, then sure you just have to find a function with equal derivatives at those points, I can't think this through completely right now but that's the gist of it.
  15. Aug 24, 2006 #14
    Can you give me an example?
  16. Aug 24, 2006 #15
    Can you give me one of what you mean by consecutive points?
  17. Aug 25, 2006 #16
    Are you referring to something like y=sin(2πx) ?
    It has equal derivatives at x=1 and x=2 (i.e., y'(1)=y'(2)=2π)
  18. Aug 25, 2006 #17
    No not quite, since that gives a nice counterexample to what I was thinking, errr... I guess not what i was thinking but what I wrote since what I was thinking was that if there are two points on a curve, if the derivative at both of these points are equal and the tangent lines at each point have the same y intercept then they must be the same line and the tangent line would be a secant line. With a sine function like the one you mention you'll have the equal derivatives and hence slopes of the tangent lines are the same but the y intercepts are different. But again with that same function consider y=1 this will be tangent to that curve for x=1/4+n where n is an integer, but this probably isn't quite what the original poster was talking about and isn't exactly what i was either. If you still don't get what I mean I'll try and write up a better post and find a good example of this tomorrow.
  19. Aug 25, 2006 #18
    :smile: Perhaps you were thinking of y=cos(2πx) ?
    Where y'(1)=y'(2)=0 and the single line y=1 is tangent to y(x) at x=1 and x=2
    (generally, as y=1 is tangent to y(x) here for all integer 'x')
    Last edited: Aug 25, 2006
  20. Aug 25, 2006 #19
    Well y=1 is tangent to sin(2πx) for all x of the form 1/4+n where n is an integer, isn't it? Because you would have sinn(π/2 + 2πn) which is 1 for every integer n, isn't it? You're example works to, and is better than mine since you don't have to have the 1/4 part.
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