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Math Help - Help in the quick way to diff y=5-X^2

  1. #1
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    Help in the quick way to diff y=5-X^2

    I need vertex,
    line of Sym x=
    Max and Min
    is the F(0)=5 a min or a max?

    Much appriciated
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  2. #2
    Super Member 11rdc11's Avatar
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    vertex is (0,5)

    line of sym is x=0

    take derative of function and equate to 0.

    y'= -2x

    0= -2x

    so x = 0 this is either a min or max. Now you can use the 2nd derative test or plug in values for the 1st derative and see where it increasing or decreasing. For example -1<0<1

    y' = -2(-1)= 2
    y' = -2(1) = -2

    so when x=0 is a max

    Hope that helps
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  3. #3
    Super Member wingless's Avatar
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    You don't need to differentiate.

    We are looking for the maximum value of 5 - x^2. It's obvious that x^2 is always positive or zero. So, the minimum of x^2 is 0. Therefore 5-x^2 is maximum at x=0.


    We can also find it using differentials. The local extrema of a function are at the roots of y' = 0. Don't forget to check for global extrema too.

    y' = -2x = 0

    x=0

    So we have a local extremum at x=0.

    We can decide whether it's maximum or minimum using the second derivative test.

    y''(0) = -2 < 0. It's a maximum.

    Don't forget to check for global maximum too. The reason for doing this is, the function may have a maximum at infinity.

    \lim_{x\to\infty} 5-x^2 = -\infty
    \lim_{x\to-\infty} 5-x^2 = -\infty

    Ok, now we're convinced that the function has a maximum only at x=0.
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