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Math Help - Crit points, extreme values, and basic algebra

  1. #1
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    Crit points, extreme values, and basic algebra

    Consider the function g(x) = x^2sqrt(5-x)

    a)find the intervals on which the function is increasing and decreasing.

    I'm stuck at basic algebra here and not sure why. This is what I've got.

    g'(x) = 2x*sqrt(5-x) - ((x^2)/(2sqrt(5-x)))

    Then I set = 0 and solve for x...

    2x*sqrt(5-x) - ((x^2)/(2sqrt(5-x))) = 0
    (x^2)/(2sqrt(5-x)) = (2x*sqrt(5-x))
    x^2 = (2x*sqrt(5-x))/(2sqrt(5-x))

    now, I'm trying to simplify by cancelation but can't remember if I can cancel out the two sqrts or the 2's or both?

    Thanks!
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  2. #2
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    Response

    Consider the function g(x) = x^2sqrt(5-x)

    a)find the intervals on which the function is increasing and decreasing.

    Well, start by taking g'(x) by using the product rule:
    g'(x) = 2xsqrt(5-x) + x^2*(1/2)(5-x)^(-1/2)*-1
    g'(x) = 2x*sqrt(5-x) -(1/2)*x^2/sqrt(5-x) = 0
    2x*sqrt(5-x) = x^2/(2*sqrt(5-x))

    Multiply both sides by 2*sqrt(5-x) to get:
    4x*(5-x) = x^2
    20x - 4x^2 = x^2
    5x^2 - 20x = 0
    5x(x-4) = 0

    x = 0, x = 4
    Hope this helps.
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  3. #3
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    Additional

    5x(x-4) = 0
    x = 0, x = 4

    Pick a point to the left of zero, say at x=-1. This causes the 5x factor to be negative, and the (x-4) factor to be negative. Their resulting product is positive, so this interval is increasing.

    Any point in between x=0 and x=4, say at x = 2 for example. This causes the 5x factor to be positive and the (x-4) factor to be negative. Their resulting product is negative, so this interval is decreasing.

    Any point to the right of x = 4, say at x = 5. This causes the 5x factor to be positive and the (x-4) factor to be positive. Their resulting product is positive, so this interval is increasing.

    The final thing looks like:

    Increasing: (-infinity, 0) U (4, infinity)
    Decreasing: (0,4)
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  4. #4
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    thanks a lot for taking time for this! Dunno why I keep getting hung up on it
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