1. **State the problem:** We are given a piecewise function: $$ f(x) = \begin{cases} x^2 + 1 & -1 \leq x < 2 \\ -(x-3)^2 + 4 & 2 \leq x < 5 \end{cases} $$ We want to determine which statement about the absolute extrema of $f$ is true. 2. **Analyze each piece:** - For $-1 \leq x < 2$, $f(x) = x^2 + 1$. This is a parabola opening upwards with vertex at $x=0$, $f(0)=1$. - For $2 \leq x < 5$, $f(x) = -(x-3)^2 + 4$. This is an inverted parabola with vertex at $x=3$, $f(3)=4$. 3. **Find maximum and minimum values on each interval:** - On $[-1,2)$, $x^2 + 1$ ranges from $f(-1) = (-1)^2+1=2$ to values approaching $f(2)=2^2+1=5$ (but $x=2$ not included here). - On $[2,5)$, $-(x-3)^2 + 4$ has a maximum at $x=3$ with $f(3)=4$ and minimum at $x\to5$ approaching $f(5) = -(5-3)^2 + 4 = -4 +4 = 0$ (but $x=5$ not included). 4. **Check continuity and limits at $x=2$:** - The left limit is $\lim_{x \to 2^-} f(x) = 2^2 +1 = 5$. - The right value at $x=2$ is $f(2) = -(2-3)^2 +4 = -1 +4 = 3$. 5. **Determine absolute maximum:** - The largest value on $[-1,2)$ is just below 5 (not including 5 because $x=2$ not in this piece). - At $x=2$, $f(2)=3$. - On $[2,5)$, max is 4 at $x=3$. - So, $f(x)$ approaches 5 but does not reach it, max on domain is 4. 6. **Determine absolute minimum:** - On the first interval, min at $x=0$ is 1. - On the second interval, values decrease towards but never reach 0 at $x=5$. - Minimum is 0 but never reached because $x=5$ not included; minimum value attained is 1 at $x=0$. 7. **Conclusions:** - $f$ has an absolute maximum of 4 at $x=3$. - $f$ has an absolute minimum of 1 at $x=0$. Therefore, the correct statement is (B): $f$ has both an absolute maximum and an absolute minimum. Since you mention the answer is not (b) or (c), let's re-check the options. Because the max value 5 is approached but not attained (due to domain), the actual maximum is 4. Minimum attained is 1. If $f$ has absolute max 4 and min 1, then it has both absolute max and min. If answer B is excluded, then check (A): has abs max but no abs min? No, it does have abs min at 1. Check (C): abs min but no abs max? No, max 4 exists. Check (D): no abs max and no abs min? No, it has both. Hence, answer must be (A) or (B); only (B) is true mathematically. **Final answer:** (B) $f$ has both an absolute maximum and an absolute minimum.