enter each answer as a whole number (like -4, 0, or 253) or dne for undefined or does not exist. ( f(6) = ) ( lim_{x o 0^-} f(x) = ) ( lim_{x o 1^+} f(x) = ) ( lim_{x o 9} f(x) = ) question help: video 1 video 2 message instructor

Question

enter each answer as a whole number (like -4, 0, or 253) or dne for undefined or does not exist. ( f(6) = ) ( lim_{x 	o 0^-} f(x) = ) ( lim_{x 	o 1^+} f(x) = ) ( lim_{x 	o 9} f(x) = ) question help: video 1 video 2 message instructor

Accepted Answer

### 1. Solve $ f(6) $
# Explanation:
## Step1: Analyze the graph at $ x = 6 $
Looking at the graph of $ y = f(x) $, we find the point corresponding to $ x = 6 $. The line segment at $ x = 6 $ is part of the line from $ x = 5 $ (where $ f(5)=2 $) to $ x = 7 $ (where $ f(7)=6 $). To find $ f(6) $, we can see the linear relationship. The slope between $ x = 5 $ and $ x = 7 $ is $ \frac{6 - 2}{7 - 5}=\frac{4}{2} = 2 $. Using the point - slope form from $ (5,2) $, the equation of the line is $ y - 2=2(x - 5) $. When $ x = 6 $, $ y-2=2(6 - 5)=2 $, so $ y=4 $. Alternatively, by directly observing the graph, the value of $ f(x) $ at $ x = 6 $ (the $ y $ - coordinate of the point on the graph at $ x = 6 $) is 4.
# Answer:
$ f(6)=\boldsymbol{4} $

### 2. Solve $ \lim_{x
ightarrow0^{-}}f(x) $
# Explanation:
## Step1: Understand the left - hand limit
The left - hand limit as $ x
ightarrow0^{-} $ means we are looking at the values of $ f(x) $ as $ x $ approaches 0 from the left side (values of $ x $ less than 0). From the graph, the left - most part of the graph (for $ x<0 $) is a line. We can see that as $ x $ approaches 0 from the left, the $ y $ - value of the function approaches 3.
# Answer:
$ \lim_{x
ightarrow0^{-}}f(x)=\boldsymbol{3} $

### 3. Solve $ \lim_{x
ightarrow1^{+}}f(x) $
# Explanation:
## Step1: Understand the right - hand limit
The right - hand limit as $ x
ightarrow1^{+} $ means we are looking at the values of $ f(x) $ as $ x $ approaches 1 from the right side (values of $ x $ greater than 1). From the graph, when $ x $ approaches 1 from the right, the function is a line segment. The point at $ x = 1 $ (right - hand side) starts the line that goes from $ (1,1) $ (the solid dot at $ x = 1 $) to $ (3,3) $? Wait, no. Wait, at $ x = 1 $, the right - hand side: the graph from $ x = 1 $ (right) goes to $ x = 3 $. The value of the function as $ x $ approaches 1 from the right: the line from $ x = 1 $ (where the solid dot is at $ y = 1 $) to $ x = 3 $ (where the open dot? No, at $ x = 1 $, the right - hand limit: looking at the graph, when $ x $ is just greater than 1, the function's value is approaching 1? Wait, no. Wait, the graph at $ x = 1 $: the solid dot is at $ (1,1) $, and the line to the right of $ x = 1 $ (for $ x>1 $) starts at $ (1,1) $ and goes up. Wait, no, the first segment: from $ x = 0 $ (left) to $ x = 1 $ (open dot at $ (1,4) $)? Wait, I think I made a mistake earlier. Let's re - examine the graph.

Looking at the graph: At $ x = 1 $, there is an open dot at $ (1,4) $ and a solid dot at $ (1,1) $. The right - hand limit as $ x
ightarrow1^{+} $ is the limit as $ x $ approaches 1 from values greater than 1. The graph for $ x>1 $ (immediately to the right of $ x = 1 $) is a line starting from the solid dot at $ (1,1) $ and going up. So as $ x $ approaches 1 from the right, the $ y $ - value approaches 1? Wait, no, that can't be. Wait, the line from $ x = 1 $ (solid dot $ (1,1) $) to $ x = 3 $ (open dot $ (3,3) $)? Wait, no, at $ x = 3 $, there is a solid dot at $ (3,2) $ and an open dot at $ (3,3) $. Wait, maybe a better way: the right - hand limit as $ x
ightarrow1^{+} $ is the value that $ f(x) $ approaches as $ x $ gets closer to 1 from the right. Looking at the graph, the part of the graph for $ x>1 $ (near $ x = 1 $) is the line that starts at $ (1,1) $ (solid dot) and goes towards $ (3,3) $ (open dot)? No, the slope between $ (1,1) $ and $ (3,3) $ is $ \frac{3 - 1}{3 - 1}=1 $. But when $ x = 1 $, the right - hand side: the function's value as $ x
ightarrow1^{+} $ is the $ y $ - value of the graph just to the right of $ x = 1 $. From the graph, the line to the right of $ x = 1 $ (for $ x>1 $) has a $ y $ - value approaching 1? Wait, no, I think I messed up the points. Let's look again:

The graph:

- For $ x $ from $ -\infty $ to $ 1 $: a line going from (let's say) $ x = 0 $ left (with $ y $ - intercept around 3) to $ x = 1 $ (open dot at $ (1,4) $).

- For $ x $ from $ 1 $ to $ 3 $: a line from $ (1,1) $ (solid dot) to $ (3,3) $ (open dot)? No, at $ x = 3 $, there is a solid dot at $ (3,2) $ and an open dot at $ (3,3) $. Wait, maybe the correct way is: when $ x
ightarrow1^{+} $, we look at the graph for $ x>1 $. The graph at $ x = 1 $ (right side) is a line that starts at $ (1,1) $ (solid dot) and goes up. So as $ x $ approaches 1 from the right, the $ y $ - value approaches 1? No, that doesn't seem right. Wait, maybe the open dot at $ (1,4) $ is the left - hand limit, and the solid dot at $ (1,1) $ is the function's value at $ x = 1 $, but the right - hand limit as $ x
ightarrow1^{+} $ is the limit of the function as $ x $ comes from the right. The line to the right of $ x = 1 $ (for $ x>1 $) is the line that goes from $ (1,1) $ to $ (3,3) $ (open dot) or $ (3,2) $ (solid dot)? I think I made a mistake in the initial analysis. Let's start over.

The right - hand limit as $ x
ightarrow1^{+} $: we look at the values of $ f(x) $ when $ x $ is slightly greater than 1. From the graph, the segment for $ x>1 $ (near $ x = 1 $) is the line that starts at $ (1,1) $ (the solid dot at $ x = 1 $) and goes towards $ (3,3) $ (the open dot at $ x = 3 $)? Wait, no, the solid dot at $ x = 3 $ is at $ y = 2 $, and the open dot is at $ y = 3 $. So the line from $ x = 1 $ (solid dot $ (1,1) $) to $ x = 3 $ (solid dot $ (3,2) $)? The slope is $ \frac{2 - 1}{3 - 1}=\frac{1}{2} $. But when $ x
ightarrow1^{+} $, the value of $ f(x) $ approaches 1? Wait, no, the solid dot at $ x = 1 $ is $ (1,1) $, so as $ x $ approaches 1 from the right, the function's value approaches 1? But that seems conflicting with the left - hand limit (which is 4). Wait, no, the left - hand limit as $ x
ightarrow1^{-} $ is 4 (since the line from the left approaches $ (1,4) $ (open dot)), and the right - hand limit as $ x
ightarrow1^{+} $ is 1 (since the line from the right approaches $ (1,1) $ (solid dot)). So $ \lim_{x
ightarrow1^{+}}f(x)=1 $? Wait, no, maybe I got the points wrong. Let's look at the graph again:

- At $ x = 1 $: open dot at $ y = 4 $ (left - hand side) and solid dot at $ y = 1 $ (right - hand side).

- The line to the right of $ x = 1 $ (for $ x>1 $) starts at $ (1,1) $ and goes up. So as $ x $ approaches 1 from the right, the $ y $ - value of the function approaches 1.

# Answer:
$ \lim_{x
ightarrow1^{+}}f(x)=\boldsymbol{1} $

### 4. Solve $ \lim_{x
ightarrow9}f(x) $
# Explanation:
## Step1: Analyze the limit as $ x
ightarrow9 $
To find $ \lim_{x
ightarrow9}f(x) $, we need to check the left - hand limit and the right - hand limit as $ x $ approaches 9.

- Left - hand limit as $ x
ightarrow9^{-} $: As $ x $ approaches 9 from the left, the graph is a line segment from $ x = 7 $ (where $ f(7)=6 $) to $ x = 9 $ (where $ f(9)=4 $ (open dot)). The value of the function as $ x $ approaches 9 from the left is 4.

- Right - hand limit as $ x
ightarrow9^{+} $: As $ x $ approaches 9 from the right, the graph is a horizontal line starting at $ x = 9 $ (open dot at $ y = 2.8 $? No, the open dot at $ x = 9 $ is at $ y = 2.8 $? Wait, no, the graph at $ x = 9 $ has an open dot at $ y = 4 $ (from the left - hand side) and an open dot at $ y = 2.8 $? No, looking at the graph, the right - hand side of $ x = 9 $ is a horizontal line with $ y = 2.8 $? Wait, no, the open dot at $ x = 9 $ (right - hand side) is at $ y = 2.8 $? No, the graph shows that at $ x = 9 $, there is an open dot at $ y = 4 $ (from the left - moving line) and an open dot at $ y = 2.8 $ (the horizontal line). Wait, no, the correct way: the left - hand limit as $ x
ightarrow9^{-} $ is the value of $ f(x) $ as $ x $ approaches 9 from the left, which is 4 (since the line from $ x = 7 $ to $ x = 9 $ has $ f(9) $ (open dot) at 4). The right - hand limit as $ x
ightarrow9^{+} $ is the value of $ f(x) $ as $ x $ approaches 9 from the right, which is 2.8? No, the graph shows a horizontal line starting at $ x = 9 $ (open dot) with $ y = 2.8 $? Wait, no, the open dot at $ x = 9 $ (right - hand side) is at $ y = 2.8 $? No, the graph has a horizontal line at $ y = 2.8 $ (around 3) starting at $ x = 9 $. Wait, the open dot at $ x = 9 $ (left - hand side) is at $ y = 4 $, and the open dot at $ x = 9 $ (right - hand side) is at $ y = 2.8 $ (close to 3). Wait, no, the limit as $ x
ightarrow9 $ exists only if the left - hand limit and the right - hand limit are equal. But from the graph, as $ x $ approaches 9 from the left, the $ y $ - value is 4, and as $ x $ approaches 9 from the right, the $ y $ - value is 3 (the open dot at $ x = 9 $ on the horizontal line is at $ y = 3 $). Wait, I think I made a mistake. Let's look at the graph again:

At $ x = 9 $, there is an open dot at $ y = 4 $ (from the line coming from $ x = 7 $) and an open dot at $ y = 3 $ (the horizontal line). So the left - hand limit $ \lim_{x
ightarrow9^{-}}f(x)=4 $ and the right - hand limit $ \lim_{x
ightarrow9^{+}}f(x)=3 $. Since $ 4
eq3 $, the limit $ \lim_{x
ightarrow9}f(x) $ does not exist. Wait, but maybe I misread the graph. Wait, the horizontal line at $ x = 9 $ (right - hand side) has an open dot at $ y = 3 $, and the line from $ x = 7 $ to $ x = 9 $ has an open dot at $ y = 4 $. So the left - hand limit is 4, the right - hand limit is 3, so the two - sided limit does not exist.

# Answer:
$ \lim_{x
ightarrow9}f(x)=\boldsymbol{DNE} $

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QUESTION IMAGE

Question

Explanation:

1. Solve \( f(6) \)

Step1: Analyze the graph at \( x = 6 \)

Step1: Understand the left - hand limit

Step1: Understand the right - hand limit

Answer:

2. Solve \( \lim_{x