What is the maximum value of f(x)= 3x^2 - x^3?

1 Answer
Eddie
Jun 24, 2016

the max value will therefore be as x \to -oo then y \to oo.

Explanation:

the dominant term in this function is the -x^3 term.

so for large negative x, the function is effectively f(-x) = - (-x)^3 = x^3. the max value will therefore be as x \to -oo then y \to oo.

equally for large positive x, the function is effectively f(x) = - (x)^3 = -x^3. the min value will therefore be as x \to +oo then y \to -oo.

maybe you are using calculus, in which case you might also be expected to look at: f'(x)= 6x - 3x^2 = x(6 - 3x)

that will be zero at critical points so x(6 - 3x) = 0 \implies x = 0, x = 2 with f(0) = 0, f(2) = 4. but these are local min and max.

you can use f''(x) = 6 - 6x to verify the nature of these turning points if that is needed eg if i have misunderstood the question. f''(0) = 6 [min], f''(2) = -6 [max]

Related questions
See all questions in Identifying Stationary Points (Critical Points) for a Function
Impact of this question
5118 views around the world
You can reuse this answer
Creative Commons License