Engineering Calculus

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f(x)=((6x-3)/(x+6)) how do you find the intervals of decreasing and increasing...I found that there were no critical points yet I was also correct that the function increased (-I,-6) and (-6,I). I was wrong that there were no decreasing intervals though.

  • Engineering Calculus -

    If fŒ(x) > 0 at each point in an interval I, then the function is said to be increasing on I.

    fŒ(x) < 0 at each point in an interval I, then the function is said to be decreasing on I.


    In your case :

    fŒ(x) = 39 /( x + 6 ) ^ 2


    39 positive

    ( x + 6 ) ^ 2 always positive except
    when x = - 6

    In point x = - 6 function has vertical asymptote


    So function :

    f( x ) = ( 6 x - 3 ) / ( x + 6)

    always increasing


    P.S.

    If you don't know how to find first derivation

    Go on:

    wolframalpha dot com

    When page be open in rectangle type:

    derivative (6x-3)/(x+6)

    and click option =

    When you see result click option:

    Show steps


    If you want to see graph of your function in google type:

    function graphs online

    When you see list of results click on:

    rechneronline.de/function-graphs

    When page be open in blue rectacangle type:

    (6x-3)/(x+6)

    Set :

    Range x-axis from -100 to 100

    Range y-axis from -100 to 100

    Then click option :

    Draw

    You will see graph of your function.

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