Find the maximum value of the function f ( x ) = x + x 2 − x 3 f(x)=x+x^2-x^3 f ( x ) = x + x 2 − x 3 for x ≥ 0 x≥0 x ≥ 0 .
Solution
Find the critical points of the function:
f ′ ( x ) = 1 + 2 x − 3 x 2 = 0 ; f'(x)=1+2x-3x^2=0; f ′ ( x ) = 1 + 2 x − 3 x 2 = 0 ;
− 3 x 2 + 2 x + 1 = 0. -3x^2+2x+1=0. − 3 x 2 + 2 x + 1 = 0. We solve the resulting equation using discriminants
D = 2 2 − 4 ⋅ ( − 3 ) ⋅ 1 = 16 ; D=2^2-4\cdot(-3)\cdot1=16; D = 2 2 − 4 ⋅ ( − 3 ) ⋅ 1 = 16 ;
x 1 = − 2 + 16 − 3 ⋅ 2 = − 1 3 ; x_1=\frac{-2+\sqrt{16}} {-3\cdot2}=-\frac1 3; x 1 = − 3 ⋅ 2 − 2 + 16 = − 3 1 ;
x 2 = − 2 − 16 − 3 ⋅ 2 = 1. x_2=\frac{-2-\sqrt{16}} {-3\cdot2}=1. x 2 = − 3 ⋅ 2 − 2 − 16 = 1. Option − 1 3 -\frac1 3 − 3 1 is not suitable, because x ≥ 0 x≥0 x ≥ 0 .
Now we define the intervals of increasing (f ′ ( x ) > 0 f'(x)>0 f ′ ( x ) > 0 ) and decreasing (f ′ ( x ) < 0 f'(x)<0 f ′ ( x ) < 0 ) functions.by the interval method.
We get that the point x = 1 x=1 x = 1 is the maximum.
f ( 1 ) = 1 + 1 2 − 1 3 = 1. f(1)=1+1^2-1^3=1. f ( 1 ) = 1 + 1 2 − 1 3 = 1. Answer
The maximum value of the function is 1.
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