The problem requires us to write a piecewise function that represents the monthly charge for electricity based on kilowatt-hours (kWh) used. The cost structure is different for customers using up to 1200 kWh and those using more than 1200 kWh.

1. Problem Description

The problem requires us to write a piecewise function that represents the monthly charge for electricity based on kilowatt-hours (kWh) used. The cost structure is different for customers using up to 1200 kWh and those using more than 1200 kWh.

2. Solution Steps

First, consider the case when $x \le 1200$. The charge is $7.10 plus 6.747 cents per kWh. We need to convert the cents to dollars, so it's$7.10 + 0.06747x$.

Next, consider the case when $x > 1200$. The charge is $88.064 plus 4.788 cents per kWh *above* 1200 kWh. Again, we convert cents to dollars, so it's$88.064 + 0.04788(x - 1200)$.

We have two equations for different usage ranges. Let's consider each of the provided answers and see if they match the problem description.

A. $f(x) = \begin{cases} 7.10 + 0.06747x & \text{if } x \le 1200 \\ 88.064 + 0.04788x & \text{if } x > 1200 \end{cases}$

In this case, for $x>1200$, it directly adds the total kWh consumed multiplied by 0.

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4

7

8

8. But the question says that only the kWh consumed above 1200 should be multiplied by 0.

0

4

7

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8. Thus, this equation is not correct.

B. $f(x) = \begin{cases} 7.10 + 0.06747x & \text{if } x \le 1200 \\ 88.064 + 0.04788(x - 1200) & \text{if } x > 1200 \end{cases}$

This option correctly describes the problem. When $x \le 1200$, the price is 7.10 plus 6.747 cents per kWh. For usage above 1200 ($x > 1200$), the price is $88.064 plus 4.788 cents for each kWh above 1200, which is represented by$x - 1200$.

C. $f(x) = \begin{cases} 7.10 + 6.747x & \text{if } x \le 1200 \\ 88.064 + 4.788x & \text{if } x > 1200 \end{cases}$

Here, we did not convert cents to dollars. This makes option C incorrect.

D. $f(x) = \begin{cases} 7.10 + 0.06747x & \text{if } x \ge 1200 \\ 88.064 + 0.04788(x - 1200) & \text{if } x < 1200 \end{cases}$

The intervals are incorrect. When $x<1200$, the first pricing scheme applies. Thus, D is wrong.

3. Final Answer

B. $f(x) = \begin{cases} 7.10 + 0.06747x & \text{if } x \le 1200 \\ 88.064 + 0.04788(x - 1200) & \text{if } x > 1200 \end{cases}$