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Calculus I - Tangent Lines

  1. Feb 23, 2009 #1
    1. The problem statement, all variables and given/known data

    A) Find all points on the graph of f(x)=x^2 with tangent lines passing through the point (5,0).

    B) Show that no line tangent to g(x)=1/x passes through the origin.

    These are problems from a lab problem solving assignment.

    2. Relevant equations


    f`(x) = (f(x+h) - f(x)) / h

    Not sure what to put, so excuse those functions if they aren't relevant.

    3. The attempt at a solution

    Well I understand what each graph looks like, but I am completely clueless as to how to approach this problem. I tried looking to my book for some sort of example, but can't find anything that directly relates to this sort of problem. I imagine the x-coordinates for problem A would have to be greater than 5. That's about as far as I can get. If I ask my professor about any of these problems she simply tells me to "think about it."

    Question B:
    It is apparent, by looking at the graph, that there is no tangent line that will pass through g(x)=1/x. I don't know how to show that though. There is a horizontal asymptote at y=0 and a vertical asymptote at x=0. So no tangent line could pass through the origin (though it would come very close). How would I show this?
  2. jcsd
  3. Feb 23, 2009 #2


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    Homework Helper

    what is the equation of a tangent line, try writing it -

    you should be able to calculate the gradient f'(a) (look at thie limit of the equation you posted for the derivative) and you know a point the line passes through (a,f(a))...
  4. Feb 23, 2009 #3


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    Science Advisor

    This is not true. The derivative is the limit of that difference quotient, as h goes to 0. But I suspect that, by the time you are calculating tangent lines, you would be expected to know that the derivative of x2 is 2x.

    You want the line to pass through (5, 0) so 0= m(5)+ b or b= -5m. You can write the equation as y= mx- 5m= m(x- 5). Now, the problem is you don't know where that line is tangent to y= x2. Let's just call that point (a, a2) so x= a. To be tangent at (a, a2) the line must pass through that point so those x and y values must satisfy the equation: a2= m(a- 5). At that point, the derivative, which is the slope of the tangent line, is 2a. Now you know you must have a2= (2a)(a- 5). Solve that equation for a and you can write out the equation of the tangent line.
    The problem said "find all points" so there may be more than one answer.

    Pretty much the same thing. Any line passing through the origin is of the form y= mx. Suppose such a line were tangent to y= 1/x at (a, 1/a). Then it must pass through that line: 1/a= ma. Further, m must be the derivative of 1/x at x=a. What is that derivative?
  5. Feb 25, 2009 #4
    I meant to put lim:x->h before the formula. Proofreading ftw.

    The way you explained it made a lot more sense. I came up with y=10x-50. This line would pass through (5,0). I don't believe there are any other tangent lines on this graph that would also pass through (5,0). I guess you could take the line tangent to y=0 as well. It passes through all points, x.

    I sort of get it. I am sure it's easier than I am making it out to be in my head. I always overcomplicate a problem. The derivative for 1/x is -1/x^2. If we are working with (a, 1/a), would the formula be 1/a = (-1/x^2)(a)+b?

    Thank you for the replies.
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