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Finding turning points on a gradient

  1. May 29, 2008 #1
    1. The problem statement, all variables and given/known data
    The gradient of the curve is:
    [tex]\frac{9-x^{2}}{(9+x^{2})^{2}}[/tex]

    Find the turning points on the curve

    2. Relevant equations



    3. The attempt at a solution

    Well for a turning point the gradient of the curve = 0
    [tex]\frac{9-x^{2}}{(9+x^{2})^{2}} = 0[/tex]

    but now what to do. in the mark scheme it seems that they have multiplied by the denomintoar, but wont that loose roots?

    Thanks
     
  2. jcsd
  3. May 29, 2008 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    No. A fraction, a/b is equal to 0 if and only if its numerator is 0. The denominator is irrelevant.

    Of course, the fraction is not defined when the denominator is 0 so you might have a critical point (max or min) there, but that is not considered a "turning point". In any case, the denominator here is (9+ x2)2 which is never 0.

    By the way, you are right that the gradient being 0 is a necessary condition for a turning point but it is not a sufficient condition. The function f(x)= x3 has f '(0)= 0 but x= 0 is not a turning point. The derivative must change signs at a turning point.
     
  4. May 30, 2008 #3
    so what your saying is that i cant get rid of the demoinator

    giving me

    x = + or - 3?

    can i just ask. what is the rules for loosing roots. when can you loose roots?

    Thanks
     
  5. May 30, 2008 #4

    exk

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    Your answer is correct. You are not losing roots. As HallsofIvy stated solving for 0 in this case involves only the numerator because a fraction is equal to 0 iff the numerator is equal to 0.
     
  6. May 30, 2008 #5

    Defennder

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    Homework Helper

    Here's an example of what it means to "lose roots" Suppose you have the equation f(x)g(x) = f(x)h(x). Note that we can cancel out f(x) on both sides. If we do that, we are left with g(x) = h(x) and we can then solve for x. However, we must also consider the possibility that some value of x exists such that f(x) = 0 and hence also satisfy f(x)g(x)=g(x)h(x). When we cancel out f(x) on both sides, we are effectively throwing away that x value, so we have to separately consider and solve for that.

    In this case you would not be discarding roots because there's no way [itex]\frac{1}{(9+x^2)^2}[/itex] can ever be zero for any value of x.
     
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