the given function $f(x)$ is defined on the interval $-9, 9$ and represented by the graph below. for the circled point, select all that apply.

graph of a function on the coordinate plane with a circled point

answer attempt 1 out of 2

- $square$ the point is an absolute maximum
- $square$ the point is an absolute minimum
- $square$ the point is a local maximum
- $square$ the point is a local minimum
- $square$ the point is neither an absolute nor a local extremum

Question

the given function $f(x)$ is defined on the interval $-9, 9$ and represented by the graph below. for the circled point, select all that apply.

graph of a function on the coordinate plane with a circled point

answer attempt 1 out of 2

- $square$ the point is an absolute maximum
- $square$ the point is an absolute minimum
- $square$ the point is a local maximum
- $square$ the point is a local minimum
- $square$ the point is neither an absolute nor a local extremum

Accepted Answer

# Brief Explanations: 1. **Absolute Minimum**: An absolute minimum on an interval $[a,b]$ is the smallest value of the function on that interval. Looking at the graph, the circled point (at $x = -6$ approximately) and other points: we check the $y$-values. The lowest $y$-value (closest to the bottom of the graph) – but wait, the circled point is at the left end, but let's analyze local and absolute extrema. - **Absolute Minimum**: The function's values: the circled point (let's see the graph) – the leftmost point (circled) and other points. Wait, the graph has a point at the bottom (the circled one? Wait, no, the $y$-axis: the circled point is at $x=-6$, $y = 0$? Wait, no, the graph: the left line comes down to the circled point (open or closed? Wait, the problem says "circled point" – looking at the graph, the circled point is at $x=-6$, and the function is defined there (closed circle? Wait, the graph: the left segment ends at the circled point (closed), then the next segment starts. Now, for absolute minimum: the absolute minimum is the smallest $f(x)$ on $[-9,9]$. Let's check the $y$-values: the bottom point (the lowest $y$) – but the circled point: is it the absolute minimum? Wait, the other points: the right side has a low point, but the circled point (left end) – let's think about local and absolute. - **Local Minimum**: A local minimum is a point where $f(x)$ is less than or equal to its neighbors. At the circled point ( $x=-6$ ), the left side (if we consider the domain $[-9,9]$, the left end is $x=-9$ (but the graph starts at $x=-9$ with a point, then comes down to $x=-6$ (circled). Wait, no, the graph: the leftmost point is at $x=-9$ (a dot), then a line to $x=-6$ (circled). Then another segment. Wait, maybe I misread. Wait, the interval is $[-9,9]$. The circled point is at $x=-6$, $y=0$ (assuming). Now, for absolute minimum: the smallest $f(x)$ on $[-9,9]$. Let's see the other points: the right side has a low point (the "valley" at $x=5$ or so, $y=3$ maybe? No, the circled point's $y$ is lower? Wait, maybe the circled point is the absolute minimum because it's the lowest $y$-value (closest to the bottom) on the entire interval. Wait, no, wait: the leftmost point ( $x=-9$ ) has a higher $y$ than the circled point ( $x=-6$ ). Then the circled point ( $x=-6$ ) is lower. Then the next segment goes up, then another peak, then a valley, then up. So the circled point ( $x=-6$ ) – is it the absolute minimum? Let's check: - Absolute minimum: the smallest $f(x)$ on $[-9,9]$. If the circled point has the lowest $y$-value (compared to all other points), then it's the absolute minimum. - Local minimum: a local minimum requires that $f(x)$ is less than or equal to its immediate neighbors. At $x=-6$, the left neighbor ( $x$ slightly less than -6, but the domain starts at $x=-9$, so the left side is the end of the interval? Wait, no, the interval is $[-9,9]$, so the left end is $x=-9$, and the circled point is at $x=-6$, which is inside the interval (not an endpoint of the entire interval, since the interval is $[-9,9]$, so $x=-6$ is interior? Wait, no, $[-9,9]$ includes all $x$ from -9 to 9. The circled point is at $x=-6$, which is between -9 and 9. Now, the function to the left of $x=-6$ (from $x=-9$ to $x=-6$) is a line decreasing to $x=-6$ (so at $x=-6$, the left neighbor ( $x$ just less than -6) has a higher $y$-value, and the right neighbor ( $x$ just more than -6) – the next segment starts, which goes up (since after $x=-6$, the graph goes up to the peak). So at $x=-6$, the function value is less than the left ( $x < -6$ ) and less than the right ( $x > -6$ )? Wait, no: the left segment is from $x=-9$ to $x=-6$, decreasing (so as $x$ increases from -9 to -6, $f(x)$ decreases). Then the next segment (from $x=-6$ to, say, $x=0$) is increasing (since it goes up to the peak). So at $x=-6$, the function value is less than the left ( $x=-9$ to $x=-6$, $f(x)$ decreases, so $f(-6)$ is less than $f(-7)$, etc.) and less than the right ( $x > -6$, $f(x)$ increases, so $f(-6)$ is less than $f(-5)$, etc.). Wait, no: if the left segment is decreasing (from $x=-9$ to $x=-6$), then $f(-9) > f(-8) > f(-7) > f(-6)$. Then the right segment (from $x=-6$ to $x=0$) is increasing, so $f(-6) < f(-5) < f(-4) < ... < f(0)$. So at $x=-6$, the function value is less than both its left (immediate left, $x$ just less than -6) and right (immediate right, $x$ just more than -6) neighbors. Wait, but for local minimum, we need a neighborhood around $x=-6$ where $f(-6)$ is the minimum. Since to the left (in $(-9, -6)$) the function is decreasing (so $f(x)$ is larger than $f(-6)$ for $x < -6$ in $(-9, -6)$), and to the right (in $(-6, 0)$) the function is increasing (so $f(x)$ is larger than $f(-6)$ for $x > -6$ in $(-6, 0)$). Therefore, $x=-6$ is a local minimum (since it's less than its immediate neighbors). But is it absolute? Let's check all points: the other low point (the "valley" at $x=5$ or so) – what's its $y$-value? The circled point ( $x=-6$ ) has $y=0$ (assuming), and the valley at $x=5$ has $y=3$ (higher). So the circled point is the lowest $y$-value on $[-9,9]$, so it's the absolute minimum. Also, since it's lower than its neighbors, it's a local minimum. Wait, but the options: let's re-examine. Wait, maybe I made a mistake. Let's recall definitions: - **Absolute minimum**: The smallest value of $f(x)$ on the entire interval $[a,b]$. - **Local minimum**: A point $c$ in $(a,b)$ (or on the interval) where $f(c) \leq f(x)$ for all $x$ in some open interval around $c$. In the graph, the circled point ( $x=-6$ ) – let's see the $y$-values. The leftmost point ( $x=-9$ ) has a higher $y$ than $x=-6$. The valley on the right ( $x=5$ ) has a higher $y$ than $x=-6$. So $f(-6)$ is the smallest on $[-9,9]$, so it's the absolute minimum. Also, since it's less than its immediate neighbors (left: $x$ just less than -6, $f(x)$ is higher; right: $x$ just more than -6, $f(x)$ is higher), it's a local minimum. Wait, but the options: the options are: - The point is an absolute maximum - The point is an absolute minimum - The point is a local maximum - The point is a local minimum - The point is neither... So let's check each: - Absolute maximum: No, the highest point is the peak near $x=0$, so no. - Absolute minimum: Yes, because it's the lowest $f(x)$ on $[-9,9]$. - Local maximum: No, it's a low point, not a maximum. - Local minimum: Yes, because in a small interval around $x=-6$, $f(-6)$ is less than $f(x)$ for $x$ near -6 (left: $x$ slightly less than -6, $f(x)$ is higher; right: $x$ slightly more than -6, $f(x)$ is higher). - Neither: No, because it's both absolute and local minimum. Wait, but maybe the circled point is at the left end? Wait, the interval is $[-9,9]$, so $x=-9$ is the left end, but the circled point is at $x=-6$, which is inside. Wait, maybe the graph's left segment is from $x=-9$ to $x=-6$ (decreasing), so at $x=-6$, it's the end of that segment, and the next segment starts. So at $x=-6$, the function value is lower than the left ( $x < -6$ ) and lower than the right ( $x > -6$ ), so it's a local minimum. And since it's the lowest on the entire interval, it's the absolute minimum. So the correct options are "The point is an absolute minimum" and "The point is a local minimum". Wait, but maybe the circled point is at the left end ( $x=-9$ ), but the graph shows a dot at $x=-9$, then a line to $x=-6$ (circled). So $x=-6$ is not the left end (the left end is $x=-9$). So $x=-6$ is an interior point (in $(-9,9)$). So as a local minimum: yes, because in a neighborhood around $x=-6$ (say, $(-7, -5)$), $f(-6)$ is less than $f(x)$ for $x$ in $(-7, -5)$ (since left of -6, $f(x)$ decreases to -6, then increases after -6). So $f(-6)$ is the minimum in $(-7, -5)$, so local minimum. And since it's the minimum on $[-9,9]$, absolute minimum. # Answer: The point is an absolute minimum, The point is a local minimum (Note: Depending on the graph's exact $y$-values, but based on the analysis, the circled point is the absolute minimum and a local minimum.) Wait, but maybe I misread the graph. Let me recheck. The graph: the leftmost point ( $x=-9$ ) has a dot, then a line to $x=-6$ (circled, closed dot). Then another curve up to a peak, then down, then a valley, then up. The circled point ( $x=-6$ ) – if its $y$-value is the lowest (e.g., $y=0$) and the valley on the right has $y=3$, then yes, absolute minimum. And since it's lower than its neighbors, local minimum. So the correct options are "The point is an absolute minimum" and "The point is a local minimum".

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