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Math Help - Simple functions - zero of f(x)

  1. #1
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    Simple functions - zero of f(x)

    I have precalculus in college. I am trying to figure out a definitive formula to answer these equations. I'm not sure if they have a formula or if you kind of just go with it. Help would be appreciated. This is a 2-step problem. (a) find the zero of the function. (b) find the min/max of the parabola.
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  2. #2
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    Quote Originally Posted by lantern4christ View Post
    I have precalculus in college. I am trying to figure out a definitive formula to answer these equations. I'm not sure if they have a formula or if you kind of just go with it. Help would be appreciated. This is a 2-step problem. (a) find the zero of the function. (b) find the min/max of the parabola.
    To find the zero of a function f(x), we simply need to solve f(x) = 0 with regards to x. That is, find x_0 such that f(x_0) = 0.

    To find the min/max of a parabola, we solve f'(x) = 0 with regards to x. Then, for those values of x we find f''(x). Say f'(x_0)=0. If f''(x_0)>0 then (x_0,f(x_0)) is a minimum point. If f''(x_0)<0, then (x_0,f(x_0)) is a maximum point.

    Example:
    Let f(x) = x^2 - 9. Find the zeroes of f(x) and find its min/max points.

    First we'll find the zeroes. Solve x^2-9=0:
    x^2-9 = 0 \Rightarrow x^2 = 9 \Rightarrow x = \pm 3
    So f(3) = f(-3) = 0.

    Now we'll find the min/max points. First we find f'(x):
    f'(x) = \frac{d}{dx}(x^2-9) = 2x

    Solve f'(x) = 0: 2x = 0 \Rightarrow x = 0
    So x=0 is a critical point. Find f''(x):
    f''(x) = \frac{d}{dx}(2x) = 2 \Rightarrow f''(0) = 2

    So (0,f(0)) = (0,-9) is a minimum point, since f''(0)>0
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  3. #3
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    Quote Originally Posted by Defunkt View Post
    To find the min/max of a parabola, we solve f'(x) = 0 with regards to x.
    Not in pre-calculus we don't, here we rely on the properties of the parabola (see this for more information).

    CB
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  4. #4
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    Oh, er, my bad then :S
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