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Derivatives and Tangents

  1. Oct 1, 2007 #1
    1. The problem statement, all variables and given/known data
    Find equations of all tangent lines to the graph of [tex]y=4x^{3}+5x-8[/tex]
    3. The attempt at a solution
    I took the derivative of the equation, which was:

    [tex]\acute{y}=12x^{2}+5[/tex]

    I remember having done these types of questions in high school, but I just can't remember, and I can't find any questions which are similar. Urg!
     
  2. jcsd
  3. Oct 1, 2007 #2
    What does a derivative give you? You may be over-thinking this.
     
  4. Oct 2, 2007 #3
    Essentially you look at the points (x,4x^3+5x-8) on the curve and for each point associate a tangent line. That is, solve (y-y(x_0))=y'(x_0)(x-x_0) for y.
     
  5. Oct 2, 2007 #4
    It gives you the slope at any single point on a line.

    I thought I should use the [tex]12x^{2}+5[/tex], and the point [tex](1,-3)[/tex] in [tex]y=mx+b[/tex], to find a [tex]b[/tex] value, but when it's all said and done I don't get a tangent line. I get [tex]y=(12x^{2}+5)x-20[/tex] which just goes through the line.

    BTW, they're looking for a total of two equations for the answer.
     
  6. Oct 2, 2007 #5

    HallsofIvy

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    It gives you the slope of the tangent line! Which is exactly what you want.
    And slope is a number not a formula in x. If you are looking for the slope of the tangent line at a point on the curve, you evaluate the derivative at the x value of that point.

    Your original question said "Find equations of all tangent lines". I THINK you are now saying "find equations for all tangent lines to 4x3+ 5x- 8 that pass through (1, -3)" but you never told us about that last part!
    One way to do that is not look at the derivative but look for solutions to the simultaneous equations y= m(x-1)+ 3 (any line through (1,3) can be written like that) and y= 4x3+ 5x- 8. For specific values of m that would give you a cubic for x which typically has three distinct answers. Look for the value of m so that equation has a double (or triple) root. That's how DesCartes found tangent lines "pre-calculus".
     
    Last edited: Oct 2, 2007
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