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

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tonys538's profile pic

tonys538 | Student, Undergraduate | (Level 1) Valedictorian

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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)

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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).

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