What are the extremes of the function 4x-8x^2?

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tonys538 | Student, Undergraduate | (Level 1) Valedictorian

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The extreme values of an expression y = f(x) lie at x = t where f'(t) = 0.

For f(x) = 4x-8x^2, f'(x) = 4 - 16x

4 - 16x = 0

=> x = 1/4

The extreme value of the function y = 4x-8x^2 is at x = 1/4.

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

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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To determine the extremes of the function, we'll have to calculate the critical values of the function, that are the roots of the first derivative of f(x).

We'll differentiate the function with respect to x, to determine the 1st derivative.

f'(x) = (4x-8x^2)'

f'(x) = 4 - 16x

We'll cancel the first derivative:

f'(x) = 0 <=> 4 - 16x = 0

We'll divide by 4:

1 - 4x = 0

We'll subtract 1:

-4x = -1

x = 1/4

Since there is a single critical value, the function has a single extreme point.

The extreme value of the function is:

f(1/4) = 4*(1/4) - 8/16

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

f(1/4) = 1/2

The extreme point of the function is represented by the pair: (1/4 ; 1/2).

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