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Some calculus exercises doubts

  1. Aug 16, 2010 #1
    1. The problem statement, all variables and given/known data

    Problem 1
    Find the equation of the tangent line on the following points of the curve.
    [itex]f(x) = \frac{1}{x - a}, a \in \mathbb{R} - \left \{ -2; 4 \} \right; x = -2, x = 4[/itex]

    Problem 2
    Find the equation of the line normal to the tangent line on the following points of the curve.
    [itex]f(x) = x^{2} - 1; x = 0[/itex]

    2. Relevant equations

    3. The attempt at a solution

    Problem 1
    [PLAIN]http://img21.imageshack.us/img21/4021/giflatex2.gif [Broken]

    As you can see my solution is completely wrong ;(



    Problem 2
    [PLAIN]http://img46.imageshack.us/img46/4265/giflatex3.gif [Broken]

    It will give a division by zero, what's going on here?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Aug 16, 2010 #2

    Mark44

    Staff: Mentor

    Evaluate your function and its derivative, and then use the values in your formulas. For #2, since f'(x) = 2x, then f'(0) = 0. This says that the tangent line is horizontal when x = 0 (i.e., at (0, 0). What does this say about the normal to the curve at this point?
     
    Last edited by a moderator: May 4, 2017
  4. Aug 16, 2010 #3

    vela

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    In problem 1, you simply forgot to substitute x=-2 in when you wrote in the expression for f'(-2) in the equation of the line.

    In problem 2, you found the slope of the tangent to be given by f'(x)=2x. At x=0, you get f'(x)=0, which means the tangent line is horizontal. The normal, therefore, will be vertical. Vertical lines do not have a finite slope, so you can't use the point-slope formula for a line.
     
  5. Aug 16, 2010 #4

    Mark44

    Staff: Mentor

    For #1, evaluate f(-2) and f'(-2) before trying to find the equation of the tangent line. My answer agrees with the book's answer that you posted.
     
  6. Aug 16, 2010 #5
    Thank you for the answers, that helped me alot, now it works... =D

    Just another question: how do I know when do I have to use the chain rule? There's any easy way of visualize it or is it just practice?
     
  7. Aug 16, 2010 #6

    Mark44

    Staff: Mentor

    Use the chain rule whenever you are differentiating a composite function. The chain rule actually comes into play in your first problem, but because the derivative of x - a is 1, the contribution from the chain rule isn't noticeable.

    If you had f(x) = 1/(2x - a) = (2x - a)-1, then f'(x) = -1(2x - a)-2 * 2 = -2/(2x - a)2. The 2 that appears in the numerator of the last expression comes from the use of the chain rule, and is the derivative of 2x - a.
     
  8. Aug 16, 2010 #7
    Thank you but I have another doubt (I think its not necessary to open another thread).

    suppose:

    [tex]f(x) = x(3x-5)[/tex]

    I need to use the chain rule here, right?

    [tex]f(x) = x u[/tex]
    [tex]u = 3x -5[/tex]
    [tex]\frac{df}{dx} = \frac{df}{du} \frac{du}{dx}[/tex]
    [tex]\frac{df}{du} = x[/tex]
    [tex]\frac{du}{dx} = 3[/tex]
    [tex]\frac{df}{dx} = 3x[/tex]

    I know that is not right, but why?
     
  9. Aug 16, 2010 #8

    Mark44

    Staff: Mentor

    You have another question, not a doubt. You have a doubt about something when you think you know what it is, but you aren't sure.

    And yes, you should start a new thread when you have a new problem.
    No. This is a product, not a composition, so you should use the product rule.
    Here it is using the product rule.
    f'(x) = x * d/dx(3x - 5) + (3x - 5) * d/dx(x)
    = x * 3 + (3x - 5) * 1
    = 3x + 3x -5 = 6x - 5.

    As a check, f(x) = 3x2 - 5x, so f'(x) = 6x - 5.
     
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