Maximum / minimum points are to be sought among critical points, i.e. the points where the first derivative y’ is zero or fails to exist.

We have:

y’ = 4 – 4(2x) = 4 – 8x = 4(1 – 2x)

So, the only critical point is x = 1/2.

By the first derivative test it is a maximum point, as y’ changes sign from “+” to “-” as x values pass through 1/2.

The same result can be obtained by the second derivative test: as y” = -8 < 0, x = 1/2 is the maximum point. ymax = 4(1/2) - 4 (1/2)^2 = 2 - 1 = 1

Dr Ivan is right, but It would be useful if you would also plot the curve so that you can understand why the formulas work. It is a very small investment in time, but it will benefit you greatly. It’s easy just to plug things into formula’s to get answers, but the people who excel in math are the ones who understand it. You could even get an A in class just by remembering formulas and knowing when to use them, so you could easily think that you are doing well. But as you advance to higher levels of math having an underlying understanding will make things easier for you and allow you to go further.