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Math plotting and simplifying

  1. May 2, 2005 #1
    [tex]0=\frac{cos^2x}{x}-\frac{1}{5}[/tex]

    [tex]\frac{1}{5}=\frac{\cos x \cos x}{x}[/tex]

    [tex]5=\frac{x}{\cos x \cos x}[/tex]

    [tex]0=x \sec ^2 x-5[/tex]

    is this true? when I plot this, it doesnt give me the same function
     
  2. jcsd
  3. May 2, 2005 #2
    I'm pretty confident that is true, but this isn't a function. You can say that [tex]0=\frac{cos^2x}{x}-\frac{1}{5}[/tex] implies and is implied by [tex]0=x \sec ^2 x-5[/tex], but that does not mean the graph of [tex]y=\frac{cos^2x}{x}-\frac{1}{5}[/tex] will be the same as the graph of [tex]y=x \sec ^2 x-5[/tex]. The two graphs cross the x-axis at the same points, i.e. they have the same roots, but otherwise they are different.
     
  4. May 2, 2005 #3
    why are they different graphs?
     
  5. May 2, 2005 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Why should they be? You did not start with a "function"- you started with an equation. Yes, it is true that [tex]0=\frac{cos^2x}{x}-\frac{1}{5}[/tex]
    is equivalent to [tex]0=x \sec ^2 x-5[/tex], but if you write functions [tex]f(x)=x \sec ^2 x-5[/tex] and [tex]g(x)=\frac{cos^2x}{x}-\frac{1}{5}[/tex], then you are only saying f(x)= 0 and g(x)= 0 have the same roots. It says nothing about other values of x.

    It is also true that x2- 1= 0 is equivalent to 1- x2= 0. Do think that means 1- x2= x2- 1 for all x?
     
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