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What values of x is there an absolute minimum for the function y= -1/3x^2 + 2x?

I found the derivative of the function which is y'=-2/3x + 2

set it to 0=y'=-2/3x + 2

What do I do now?

  • Calculus -

    what is the second derivative of y?

    y"=-2/3, which means that the curve is downward, so that is a maximum...

    The curve is a parabola, with a max at (x=3), and has no absolute min.

  • Calculus -

    Thank you!

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