Question

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

Answer 1

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]`