Answer to Question #94897 in Calculus for Sujon Rana

Question

Find the maximun value of the function f(x)=x+x^2-x^3 for x>=0

Answer 1

First of all, let's find derivative of our function:

$(x+x^2-x^3)'=1+2x-3x^2=0$

$D=4-4*(-3)=4^2, \\ x_1= \cfrac{-2-4}{-6}=1, x_2=\cfrac{-2+4}{-6}=-\cfrac{1}{3},$

$f'(1-\epsilon)>0, f'(1+\epsilon)<0 \implies x=1 -$ local maximum, now let's prove that it's maximum value of the function when $x\in[0,+\infin): f(0)=0<f(1), f''(x)=2-6x \implies \forall x \in [1, +\infin): f''(x)<0,$ for this reason:

$f'(x)$ is monotonically decreasing function  $\forall x \in [1, +\infin)$ $\implies \forall x>1: f'(x)<0 \implies f(x) -$ monotonically decreasing function. $f(1)=f_{max}(x)$ for $x\in[0, +\infin)$