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    y=x^(3/2)+48/x, x>0
    Find the value of x and the value of y when dy/dx=0
    Could you show how you got the answer as well please?

    Thank you!!
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    How can you express \frac{48}{x} as something you can differentiate?

    Once you've differentiated it, you can substitute your value of \frac{dy}{dx} into the equation to find x, and use your original equation (y=) to find y.
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    (Original post by 0katee)
    y=x^(3/2)+48/x, x>0
    Find the value of x and the value of y when dy/dx=0
    Could you show how you got the answer as well please?

    Thank you!!
    Ok since you asked for it... even though you're not supposed to give full solutions
    Spoiler:
    Show
    y=x^{\frac{3}{2}} +48x^{-1}

    \dfrac{3}{2} x^{\frac{1}{2}} - 48x^{-2}
 
 
 
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