Question:

Step 1: Analyze the function on each piece. The function is defined as: $$f(x) = \begin{cases} -x^3 + a, & 0 \leq x < 1 \\ x, & 1 \leq x \leq 3 \end{cases}$$ For the interval $0 \leq x < 1$: $f(x) = -x^3 + a$. The derivative $f'(x) = -3x^2$, which is negative for $x \in (0,1)$. Thus, $f(x)$ is a decreasing function on this interval. At $x=0$, $f(0) = a$. As $x \to 1^-$, $f(x) \to -1^3 + a = -1+a$. For the interval $1 \leq x \leq 3$: $f(x) = x$. The derivative $f'(x) = 1$, which is positive. Thus, $f(x)$ is an increasing function on this interval. At $x=1$, $f(1) = 1$. At $x=3$, $f(3) = 3$. Step 2: Condition for an absolute minimum at $x=1$. For $f(x)$ to have an absolute minimum at $x=1$, we must have $f(x) \geq f(1)$ for all $x \in [0,3]$. From the definition, $f(1) = 1$. For $x \in [1,3]$, $f(x) = x$. Since $x \geq 1$ for this interval, the condition $f(x) \geq f(1)$ is satisfied. For $x \in [0,1)$, $f(x) = -x^3 + a$. We require $-x^3 + a \geq 1$. This implies $a \geq 1 + x^3$. This inequality must hold for all $x \in [0,1)$. The function $g(x) = 1+x^3$ is increasing on $[0,1)$, so its supremum on this interval is $\lim_{x \to 1^-} (1+x^3) = 1+1^3 = 2$. Therefore, we must have $a \geq 2$. The minimum positive integral value of $a$ for $f(x)$ to have an absolute minimum at $x=1$ is $2$. Step 3: Condition for an absolute maximum at $x=3$. For $f(x)$ to have an absolute maximum at $x=3$, we must have $f(x) \leq f(3)$ for all $x \in [0,3]$. From the definition, $f(3) = 3$. For $x \in [1,3]$, $f(x) = x$. Since $x \leq 3$ for this interval, the condition $f(x) \leq f(3)$ is satisfied. For $x \in [0,1)$, $f(x) = -x^3 + a$. We require $-x^3 + a \leq 3$. This implies $a \leq 3 + x^3$. This inequality must hold for all $x \in [0,1)$. The function $h(x) = 3+x^3$ is increasing on $[0,1)$, so its infimum on this interval is $3+0^3 = 3$. Therefore, we must have $a \leq 3$. The maximum positive integral value of $a$ for $f(x)$ to have an absolute maximum at $x=3$ is $3$. Step 4: Condition for an absolute maximum at $x=0$. For $f(x)$ to have an absolute maximum at $x=0$, we must have $f(x) \leq f(0)$ for all $x \in [0,3]$. From the definition, $f(0) = a$. For $x \in [0,1)$, $f(x) = -x^3 + a$. We require $-x^3 + a \leq a$. This simplifies to $-x^3 \leq 0$, which is true for all $x \in [0,1)$. So this part of the condition is always satisfied. For $x \in [1,3]$, $f(x) = x$. We require $x \leq a$. This inequality must hold for all $x \in [1,3]$. The maximum value of $x$ on this interval is $3$. Therefore, we must have $a \geq 3$. The minimum positive integral value of $a$ for $f(x)$ to have an absolute maximum at $x=0$ is $3$.