Answer to Question #94176 in Algebra for Yordanov

Question

Find the maximum value of the function f(x) = x + x'2 − x'3
for x ≥ 0

Answer 1

Find the maximum value of the function $f(x)=x+x^2-x^3$ for $x≥0$.

Solution

Find the critical points of the function:

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

$$-3x^2+2x+1=0.$$

We solve the resulting equation using discriminants

$$D=2^2-4\cdot(-3)\cdot1=16;$$

$$x_1=\frac{-2+\sqrt{16}} {-3\cdot2}=-\frac1 3;$$

$$x_2=\frac{-2-\sqrt{16}} {-3\cdot2}=1.$$

Option $-\frac1 3$ is not suitable, because $x≥0$.

Now we define the intervals of increasing ($f'(x)>0$) and decreasing ($f'(x)<0$) functions.by the interval method.

We get that the point $x=1$ is the maximum.

$$f(1)=1+1^2-1^3=1.$$

Answer

The maximum value of the function is 1.