Question

The maximum value of the function $y = x(x-1)^2, 0 \le x \le 2$ is

• A. 0
• B. $\frac{4}{27}$
• C. -4
• D. none

Answer 1

Answer: (D)

D

Answer 2

To find the maximum value of the function $y = x(x-1)^2$ in the interval $0 \le x \le 2$, we follow these steps:

1. Expand the function: $y = x(x^2 - 2x + 1) = x^3 - 2x^2 + x$

2. Find the first derivative of the function: $f'(x) = \frac{d}{dx}(x^3 - 2x^2 + x) = 3x^2 - 4x + 1$

3. Find the critical points by setting the first derivative to zero: $3x^2 - 4x + 1 = 0$ This is a quadratic equation. We can factor it: $3x^2 - 3x - x + 1 = 0$ $3x(x - 1) - 1(x - 1) = 0$ $(3x - 1)(x - 1) = 0$ This gives two critical points: $3x - 1 = 0 \implies x = \frac{1}{3}$ $x - 1 = 0 \implies x = 1$

4. Check if the critical points lie within the given interval $0 \le x \le 2$: Both $x = \frac{1}{3}$ and $x = 1$ are within the interval $[0, 2]$.

5. Evaluate the function at the critical points and the endpoints of the interval: The endpoints of the interval are $x = 0$ and $x = 2$.

   • At $x = 0$ (endpoint): $f(0) = 0(0 - 1)^2 = 0(1) = 0$

   • At $x = \frac{1}{3}$ (critical point): $f\left(\frac{1}{3}\right) = \frac{1}{3}\left(\frac{1}{3} - 1\right)^2 = \frac{1}{3}\left(-\frac{2}{3}\right)^2 = \frac{1}{3}\left(\frac{4}{9}\right) = \frac{4}{27}$

   • At $x = 1$ (critical point): $f(1) = 1(1 - 1)^2 = 1(0)^2 = 0$

   • At $x = 2$ (endpoint): $f(2) = 2(2 - 1)^2 = 2(1)^2 = 2(1) = 2$

6. Compare all the evaluated values to find the maximum: The values are $0, \frac{4}{27}, 0, 2$.

   Comparing $0, \frac{4}{27}, 2$, the maximum value is $2$.

Since $2$ is not among options A, B, or C, the correct option is D.