Show that the maximum and minimum values of y occurs when x^3=8+- 2(14)^1/2
Right.As I know at stationary point,dy/dx=0 so from dy/dx=(2y-x^2)/(y^2-2x),from there I can find y=(x^2)/2
Ok that will fetch you the maximum and minimum of y.then I substitute x=2+2(14)^(1/2) and x=2-2(14)^(1/2) to get y values right?
I'm not sure if I understood this one.Then I reciprocate the dy/dx =dx/dy and then I do the sign test? But I got all + signs which means I fail to prove y has maximum and minimum values.
Ok that will fetch you the maximum and minimum of y.
I'm not sure if I understood this one.
But as soon as you get the ##y=x^2/2##, can you use it and the 'given' to solve for values of x?
How did you get this? Probably by the previous x^3 equation which they have given?I have used the calculator to change the x^3 values into 15.84 and -5.84
but when I substitute the x values and y values into my second derivative but in the end I got all positive value which means only local minimum values only.