Let f(x) = x^3, x in [-1, 1]. Then which of t | vedclass

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(C, D) Given the function $f(x) = x^3$ on the interval $x \in [-1, 1]$.$1$. Check for extrema: $f'(x) = 3x^2$. Setting $f'(x) = 0$ gives $x = 0$. Sinc

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$f$ is bounded on $[-1, 1]$

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(C, D) Given the function $f(x) = x^3$ on the interval $x \in [-1, 1]$.$1$. Check for extrema: $f'(x) = 3x^2$. Setting $f'(x) = 0$ gives $x = 0$. Since $f'(x) \ge 0$ for all $x \in [-1, 1]$, the function is strictly increasing. Thus, it has no local minimum or maximum at $x = 0$. The absolute minimum is at $x = -1$ $(f(-1) = -1)$ and the absolute maximum is at $x = 1$ $(f(1) = 1)$. So, options $A$ and $B$ are incorrect.$2$. Continuity: The function $f(x) = x^3$ is a polynomial function, which is continuous for all real numbers, including the interval $[-1, 1]$. Thus, option $C$ is correct.$3$. Boundedness: Since $f(x)$ is continuous on the closed interval $[-1, 1]$, by the Extreme Value Theorem, it is bounded. Specifically, $-1 \le f(x) \le 1$ for all $x \in [-1, 1]$. Thus, option $D$ is correct.Therefore, the correct statements are $C$ and $D$.

Let $f(x) = x^3$, $x \in [-1, 1]$. Then which of the following are correct?

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