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Math Help - decreasing intervals

  1. #1
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    decreasing intervals

    Find all open intervals on which the function f(x) = \frac {x^2} {x^2 + 4} is decreasing
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Samantha View Post
    Find all open intervals on which the function f(x) = \frac {x^2} {x^2 + 4} is decreasing
    That is find the open intervals on which \frac{d}{dx}\left[\frac{x^2}{x^2+4}\right] \le 0

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by CaptainBlack View Post
    That is find the open intervals on which \frac{d}{dx}\left[\frac{x^2}{x^2+4}\right] \le 0

    RonL
    Now:

    <br />
\frac{d}{dx}\left[\frac{x^2}{x^2+4}\right]=\frac{2x}{x^2+4}-\frac{2x^3}{(x^2+4)^2}<br />

    which is less than or equal to zero when:

    <br />
2x(x^2+4)-2x^3 \le 0<br />

    RonL
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  4. #4
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    Hello, Samantha!

    Exactly where is your difficulty?
    . . You don't know the Quotient Rule?
    . . You don't understand increasing/decreasing?
    . . You can't handle inqualities?


    Find all open intervals on which the function f(x) \:=\: \frac {x^2} {x^2 + 4} is decreasing
    Quotient Rule: . f'(x) \;=\;\frac{(x^2+4)\cdot2x - x^2\cdot2x}{(x^2+4)^2}\;=\;\frac{8x}{(x^2+4)^2}


    The function is decreasing where f'(x) \,<\,0

    . . So we have: . \frac{8x}{(x^2+4)^2} \:<\:0

    Since the denominator is always positive, we have: . 8x \:<\:0\quad\Rightarrow\quad x \:<\:0


    Therefore, f(x) is decreasing on the interval: . (-\infty,\:0)

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  5. #5
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    Quote Originally Posted by Samantha View Post
    Find all open intervals on which the function f(x) = \frac {x^2} {x^2 + 4} is decreasing
    The curve also shows that f(x) is decreasing on (-infinity,0)
    Attached Thumbnails Attached Thumbnails decreasing intervals-july57.gif  
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