First-class postage is $$ \$ 0.42 $$ for the first. ounce (or any fraction thereof) and $$ \$ 0.15 $$ for each additional ounce (or fraction thereof) up to a maximum weight of 3.5 ounces. (A) Write a piecewise definition of the first-class postage $$ P(x) $$ for a letter weighing $$ x $$ ounces. (B) Graph $$ P(x) $$ for $$ 0

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### (A) Piecewise Definition of the First-Class Postage $$ P(x) $$

The postage cost is defined based on the weight of the letter $$ x $$ ounces. The pricing structure is as follows:

1. For $$ 0 < x \leq 1 $$: The cost is $$ \$0.42 $$.
2. For $$ 1 < x \leq 2 $$: The cost is $$ \$0.42 + \$0.15 $$ for the second ounce, which totals $$ \$0.57 $$.
3. For $$ 2 < x \leq 3 $$: The cost is $$ \$0.42 + 2 \times \$0.15 $$ for the third ounce, which totals $$ \$0.72 $$.
4. For $$ 3 < x \leq 3.5 $$: The cost is $$ \$0.42 + 3 \times \$0.15 $$ for the maximum weight, which totals $$ \$0.87 $$.

Thus, the piecewise function $$ P(x) $$ can be defined as:

$$
P(x) = 
\begin{cases} 
0.42 & \text{if } 0 < x \leq 1 \
0.57 & \text{if } 1 < x \leq 2 \
0.72 & \text{if } 2 < x \leq 3 \
0.87 & \text{if } 3 < x \leq 3.5 
\end{cases}
$$

### (B) Graph $$ P(x) $$ for $$ 0 < x \leq 3.5 $$

To graph $$ P(x) $$, we will plot the function based on the piecewise definition. The function will have constant values in the specified intervals:

- From $$ 0 < x \leq 1 $$, $$ P(x) = 0.42 $$
- From $$ 1 < x \leq 2 $$, $$ P(x) = 0.57 $$
- From $$ 2 < x \leq 3 $$, $$ P(x) = 0.72 $$
- From $$ 3 < x \leq 3.5 $$, $$ P(x) = 0.87 $$

I will now create the graph for $$ P(x) $$.

### (C) Continuity of $$ P(x) $$ at $$ x=2.5 $$ and $$ x=3 $$

To determine the continuity of $$ P(x) $$ at these points, we need to check the following:

1. **At $$ x = 2.5 $$**:
   - $$ P(2.5) = 0.72 $$ (since $$ 2 < 2.5 \leq 3 $$)
   - $$ \lim_{x \to 2.5^-} P(x) = 0.72 $$ (approaching from the left)
   - $$ \lim_{x \to 2.5^+} P(x) = 0.72 $$ (approaching from the right)

Since both limits exist and equal $$ P(2.5) $$, we conclude that $$ P(x) $$ is continuous at $$ x = 2.5 $$.

2. **At $$ x = 3 $$**:
   - $$ P(3) = 0.72 $$ (since $$ 2 < 3 \leq 3 $$)
   - $$ \lim_{x \to 3^-} P(x) = 0.72 $$ (approaching from the left)
   - $$ \lim_{x \to 3^+} P(x) = 0.87 $$ (approaching from the right)

Since the left limit does not equal the right limit, $$ P(x) $$ is not continuous at $$ x = 3 $$.

### Summary of Continuity

- At $$ x = 2.5 $$: Yes, $$ P(x) $$ is continuous.
- At $$ x = 3 $$: No, $$ P(x) $$ is not continuous.

### Final Answer for Continuity at $$ x = 2.5 $$

The correct explanation for continuity at $$ x = 2.5 $$ is:
- Yes, because $$ \lim_{x \rightarrow 2.5} P(x) $$ exists, $$ P(2.5) $$ exists, and $$ \lim_{x \rightarrow 2.5} P(x) = P(2.5) $$.
$$ P(x) $$ is continuous at $$ x = 2.5 $$ because the left and right limits exist and are equal to $$ P(2.5) $$.

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