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Derivatives - Finding Tangent

  1. Mar 21, 2006 #1
    Question: Find an equation of the straight line that is tangent to the graph of f(x) = √x+1 and parallel to x-6y+4=0.

    I've found f`(x).
    When I write lim, assume "h->0" (as h approaches 0) is beneath it. I'd like to use LaTex to show you, but it seems too confusing to figure out.
    f`(x)=lim f(x+h) - f(x) / h
    f`(x)=lim √x+h+1 - √x+1 / h x √x+h+1 + √x+1 / √x+h+1 + √x+1
    f`(x)=lim x+h+1-x-1 / h(√x+h+1 + √x+1)
    f`(x)=lim 1 / h(√x+h+1 + √x+1)
    f`(x)= 1 / 2(√x+1)

    So that's f`(x). The problem is, I'm not sure where to go from here. I'm trying to find the equation of the line, and it has to be parallel to x-6y+4=0. Usually when I do a question like this, I'm given the point, but what do you do when you're given a line parallel to the tangent?
  2. jcsd
  3. Mar 21, 2006 #2


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    Try re-arranging [itex] x - 6y + 4 = 0[/itex] into the form [itex]y = mx + c[/itex]. Does that give you any ideas? :smile:
  4. Mar 21, 2006 #3
    You did the most difficult, i.e, find f'(x).
    Now, you should find the point of the function where the derivative is equal to the slope of the given straight line.
    But first we need to find the slope or gradient of the straight line.

    x - 6y + 4 = 0 <=>
    y = x/6 + 2/3

    m = 1/6, because the equation of a straight line is y = mx + c (the constant m is often called the slope or gradient while c is the y-intercept)*

    Then, we resolve the following equation:

    1/6 = 1 / 2(√x+1)

    I won't resolve it, since you have shown that you are able to do it. If you have difficulties resort to the calculator and intersect the functions y = 1 / 2(√x+1) and y = 1/6.

    Hope I could help.

    * To remind:
    See- http://en.wikipedia.org/wiki/Linear_function
    Last edited by a moderator: Mar 21, 2006
  5. Mar 21, 2006 #4


    In the interest of effective communication, I highly recommend the use of grouping symblos. The derivative of f(x) = √x+1, for instance, is not 1 / 2(√x+1). Rather, f '(x) = 1/(2√(x)). That said, I assume that you mean f(x) =√(x+1), in which case f '(x) = 1 / [2√(x+1)]. Now set this expression equal to 1/6 and you are in business. As you have it, i.e., 1/6 = 1 / 2(√x+1), the variable stands alone within the radical. Moreover, you have expressed (√x+1) as a numerator rather than a denominator. By bracketing as [2(√x+1)], the reader understands your intent to place the entire expression beneath the fraction bar.


    Rich B.
  6. Mar 21, 2006 #5
    Aha, that makes perfect sense. I probably should have thought it out more. I just worked out the answer using your tips and it shows to be correct.
    Thanks for the help guys, and I'll keep that in mind Rich. :)
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