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.
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
x=3
What do I do now?
2 answers
Thank you!
1 answer