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

Jun 24, 2016

the max value will therefore be as $x \setminus \to - \infty$ then $y \setminus \to \infty$.

Explanation:

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

so for large negative x, the function is effectively $f \left(- x\right) = - {\left(- x\right)}^{3} = {x}^{3}$. the max value will therefore be as $x \setminus \to - \infty$ then $y \setminus \to \infty$.

equally for large positive x, the function is effectively $f \left(x\right) = - {\left(x\right)}^{3} = - {x}^{3}$. the min value will therefore be as $x \setminus \to + \infty$ then $y \setminus \to - \infty$.

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

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

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