Answer to Question #100205 in Differential Equations for Beast

Question

Determine the maximum and minimum of the function f (x) = -x3 + 2x2 – x [0.2].

Answer 1

$1) We \, find \, f'(x):\\ f'(x) = -3x^2 + 4x-1; \\ 2) f'(x) = 0 \\ -3x^2 + 4x - 1 = 0; \\ 3x^2 - 4x +1 = 0; \\ D = b^2 - 4ac = (-4)^2 - 4*3*1 = 16 - 12 = 4; \\ x_1 = \frac{-b - \sqrt{D}}{2a} = \frac{4 - 2}{2*3} = \frac{2}{6} = \frac{1}{3}; \\ x_2 = \frac{-b + \sqrt{D}}{2a} = \frac{4 + 2}{2*3} = \frac{6}{6} = 1;$

3)

$4) \newline f(0) = -0^3+2*0^2-0= 0; \newline f(\frac{1}{3}) = -(\frac{1}{3})^3 + 2(\frac{1}{3})^2 - \frac{1}{3} = -\frac{4}{27};\newline f(1) = -1^3 +2*1^2-1 = 0; \newline f(2) = -2^3 + 2*2^2 - 2 = -2;\\ max \, f=max \{0; -\frac{4}{27};-2\}=0;\\ min \,f=min \{0; -\frac{4}{27};-2\}=-2.\\$

5) Therefore, the function  $f(x) = -x^3 + 2x^2 - x$ in the interval $[0,2]$  has its own minimum equal to $-2$ (the function attains this minimum at the point $x=2$) and the function has the maximum equal to 0 (the function attains this maximum at the points $x = 0$ and $x=1$).