Function Analysis: $f(x) = x^3 - 3x^2 + 2$ on $(-1, 3]$

We analyze the function $f(x) = x^3 - 3x^2 + 2$ over the interval $(-1, 3]$ to determine the correctness of the given statements.

Derivative Analysis for Extrema

First, find the critical points by calculating the first derivative and setting it to zero.

• First derivative: $f'(x) = \frac{d}{dx}(x^3 - 3x^2 + 2) = 3x^2 - 6x$.
• Set $f'(x) = 0$: $3x^2 - 6x = 0 \implies 3x(x - 2) = 0$.
• The critical points are $x = 0$ and $x = 2$. Both lie within the domain $(-1, 3]$.

Second Derivative Test

Use the second derivative test to classify the critical points.

• Second derivative: $f''(x) = \frac{d}{dx}(3x^2 - 6x) = 6x - 6$.
• At $x = 0$: $f''(0) = 6(0) - 6 = -6$. Since $f''(0) < 0$, $x = 0$ corresponds to a local maximum.
• At $x = 2$: $f''(2) = 6(2) - 6 = 6$. Since $f''(2) > 0$, $x = 2$ corresponds to a local minimum.

Function Values and Interval Behavior

Evaluate the function at critical points and consider the interval endpoints.

• Local maximum value: $f(0) = 0^3 - 3(0)^2 + 2 = 2$.
• Local minimum value: $f(2) = 2^3 - 3(2)^2 + 2 = 8 - 12 + 2 = -2$.
• Limit at the left boundary: As $x \to -1^+$, $f(x) \to (-1)^3 - 3(-1)^2 + 2 = -1 - 3 + 2 = -2$.
• Value at the right boundary: $f(3) = 3^3 - 3(3)^2 + 2 = 27 - 27 + 2 = 2$.

Analysis of Options

• Option 1: $f(x)$ has exactly two roots in $[-0.9, 0]$.
  On the interval $[-0.9, 0]$, $f'(x) = 3x(x-2)$ is positive (since $x<0$ and $x-2<0$). Thus, $f(x)$ is strictly increasing. $f(-0.9) = (-0.9)^3 - 3(-0.9)^2 + 2 = -0.729 - 2.43 + 2 = -1.159$. $f(0) = 2$. Since the function increases from a negative value to a positive value, it crosses the x-axis exactly once. This statement is incorrect.
• Option 2: $f(x)$ has a minimum at 2 only.
  $x = 2$ is a local minimum, and $f(2) = -2$. The minimum value of the function on the interval $(-1, 3]$ is $-2$. This minimum value is attained at $x = 2$. Although $\lim_{x\to -1^+} f(x) = -2$, the minimum value is uniquely achieved at the point $x=2$ within the specified interval context. This statement is correct.
• Option 3: $f(x)$ has maximum at 0 only.
  $x = 0$ is a local maximum ($f(0) = 2$). However, $f(3) = 2$. The maximum value of $f(x)$ on $(-1, 3]$ is $2$, which occurs at both $x = 0$ and $x = 3$. This statement is incorrect.
• Option 4: $f(x)$ has a root at 1.
  Evaluate $f(1)$: $f(1) = (1)^3 - 3(1)^2 + 2 = 1 - 3 + 2 = 0$. Since $f(1) = 0$, $x=1$ is a root. This statement is correct.

Conclusion

The correct statements are Option 2 and Option 4.