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

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The extreme values of a function y = f(x) lie at the point where y' = f'(x) = 0

For y=4x-16x^2, y' = 4 - 32x

4 - 32x = 0

4 = 32x

x = 1/8

At x = 1/8, y = 4*(1/8) - 16*(1/8)^2 = 1/2 - 1/4 = 1/4

The extreme point of the function y = 4x - 16x^2 lies at (1/8, 1/4)

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).**