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Find the equation of a cubic graph

  1. Jul 16, 2017 #1
    1. The problem statement, all variables and given/known data
    ##x^3 - 4x^2 + ax + b##
    tangent to x-axis at x = 3

    2. Relevant equations


    3. The attempt at a solution
    if the graph tangent at x = 3, means at x =3, y = 0
    my questions is, is at x = 3 the graph's gradient (slope) = 0 ?

    if yes why?

    if yes then means dy/dx = 0
    ##3x^2 - 8x + a = 0##
    ##3.3^2 - 8.3 + a = 0##
    ##27 - 24 + a = 0##
    a = -3

    at x=3, y = 0
    ##3^3 - 4.3^2 - 3.3 + b = 0##
    ##27 - 39 - 9 + b = 0##
    ## b = 18 ##

    so f(x) = ## x^3 -4x^2 -3x + 18 ##

    2nd question :
    " find all x such that f(x) has points in common with the x-axis? "
    what does it mean by "points in common with the x-axis"?
     
  2. jcsd
  3. Jul 16, 2017 #2

    stevendaryl

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    Staff Emeritus
    Science Advisor

    Yes, if a graph [itex]y = f(x)[/itex] is tangent to the x-axis at point [itex]x_0[/itex], that means that [itex]\frac{df}{dx} = 0[/itex] at [itex]x_0[/itex]. That can be taken as the definition of "tangent": Two curves [itex]f(x)[/itex] and [itex]g(x)[/itex] are tangent at point [itex]x_0[/itex] if they have the same linear approximation at that point:

    [itex]f(x_0) = g(x_0)[/itex]
    [itex]\frac{df}{dx}|_{x=x_0} = \frac{dg}{dx}|_{x=x_0}[/itex]

    If the curve [itex]y = g(x)[/itex] is the x-axis, then [itex]g(x) = \frac{dg}{dx} = 0[/itex]

    Well, the x-axis is the set of points [itex](x,y)[/itex] such that [itex]y=0[/itex]. A graph [itex]y=f(x)[/itex] is the set of points [itex](x,y)[/itex] such that [itex]f(x) = y[/itex]. So the points in common between the x-axis and that graph are the sets of all points [itex](x,0)[/itex] such that [itex]f(x) = 0[/itex].
     
  4. Jul 17, 2017 #3

    epenguin

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    Homework Helper
    Gold Member

    Just think - if the slope at this point (3, 0) were not zero then the curve is pointing upwards near this point if the slope is positive, or downwards if it is negative. In either case it goes from under the x axis to over it - it is not tangent there. They have given you an example which makes it easy to see, because it is tangent actually on the x axis. But general you will get problems with minima and maxima that are not on the x-axis, just imagine a different horizontal line(y = C) instead of the x-axis in the minimum or maximum is tangent to that.

    If you think back you were probably explained this in class - get it clear because it is going to come in all the time in calculus.
     
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