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    y= 24 ln x -x^3 + 8

    find the co-ordinates of the stationary point?

    thanks in adv
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    (Original post by asem93)
    y= 24 ln x -x^3 + 8

    find the co-ordinates of the stationary point?

    thanks in adv
    What is the derivative of 24lnx? What is the derivative of -x^3? What is the derivative of 8? All with respect to x.
    Add these three results together to get dy/dx. Set the resultant expression equal to zero and solve for x. Then sub in this x-value into your equation for y and find y.
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    (Original post by Farhan.Hanif93)
    What is the derivative of 24lnx? What is the derivative of -x^3? What is the derivative of 8? All with respect to x.
    Add these three results together to get dy/dx. Set the resultant expression equal to zero and solve for x. Then sub in this x-value into your equation for y and find y.
    so dy/dx = 24/x -3x^2 ??

    therefore 24/x - 3x^2 = 0
    x = 2

    am I right .... y= 24 ln 2 -2^3 + 8 = 16.6355


    Then they ask me determine whether it's max/min point??

    Thanks, Hanif93 I voted you as the best mathematician in TSR
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    (Original post by asem93)
    so dy/dx = 24/x -3x^2 ??

    therefore 24/x - 3x^2 = 0
    x = 2

    am I right .... y= 24 ln 2 -2^3 + 8 = 16.6355


    Then they ask me determine whether it's max/min point??

    Thanks, Hanif93 I voted you as the best mathematician in TSR
    You can leave y in an exact form i.e. 24ln2 will do.
    If you want to determine the nature of the stationary point, differentiate dy/dx and plug in x=2. If the result is greater than zero, it is a minimum. If it's less than zero, it's a maximum.
    And thanks, I guess.
 
 
 
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