Find the extreme values of function y=4x-16x^2?

There is one extreme value of thegiven
function.

To determine it, we'll have to calculate the
critical value of the function, that is the root of the first
derivative.

f'(x) = -32x +
4

We'll cancel f'(x):

f'(x) =
0

-32x + 4 = 0

-32x =
-4

32x = 4

x =
4/32

x = 1/8

Since there is
one critical value, it means that the function has one extreme value and it is the
maximum point of the function.

f(1/8) = 4/8 - 16/64 = 1/2 -
4/16

f(1/8) = 1/2 - 1/4

f(1/8)
= 1/4

The maximum value of the function is
represented by the pair: (1/8 ; 1/4).