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The question asks to identify which of the three given functions has a **jump discontinuity at \( x = 2 \)** and a **removable discontinuity at \( x = 4 \)**, while being continuous elsewhere.

Let's examine the functions one by one:

### Option (a):
\[ f(x) = \frac{2}{(x-2)(x-4)} \]

- This function has **potential discontinuities at \( x = 2 \)** and \( x = 4 \)** because those values make the denominator zero.
- The discontinuity at \( x = 2 \) is a **non-removable discontinuity** (likely a vertical asymptote, since the numerator remains 2).
- The discontinuity at \( x = 4 \) is **removable** because the denominator can be factored to \( (x-4) \), and thus can be simplified.
  
This matches the condition of having a **removable discontinuity at \( x = 4 \)** and a **jump discontinuity at \( x = 2 \)**.

### Option (b):
\[ f(x) = \begin{cases} 
1 & \text{if } x \leq 2 \\
x - 3 & \text{if } 2 < x < 4 \text{ or } x > 4 \\
3 & \text{if } x = 4 
\end{cases} \]

- At \( x = 2 \), the function is continuous because it is defined as \( f(x) = 1 \) for \( x \leq 2 \), and it smoothly connects for values larger than 2 (since \( x-3 \) is continuous in its interval).
- At \( x = 4 \), the function jumps from \( f(x) = 1 \) to \( f(x) = 3 \), indicating a **jump discontinuity**.

This doesn't match the condition because there's no removable discontinuity at \( x = 4 \).

### Option (c):
\[ f(x) = \begin{cases} 
2 - x^2 & \text{if } x \leq 2 \\
\frac{1}{x^2 - 4x} & \text{if } x > 2 
\end{cases} \]

- The function has a **potential discontinuity** at \( x = 2 \), but since the left-hand limit and right-hand limit do not match, it indicates a **jump discontinuity** at \( x = 2 \).
- The denominator \( x^2 - 4x = x(x - 4) \) suggests a discontinuity at \( x = 4 \). The discontinuity at \( x = 4 \) is **non-removable** since it's in the denominator.

This option does not satisfy the condition either.

### Conclusion:
The correct answer is **(a)** because it has a **jump discontinuity at \( x = 2 \)** and a **removable discontinuity at \( x = 4 \)**, while being continuous elsewhere.

Would you like further details on this solution? 

Here are some related questions to explore:

1. What is the difference between removable and non-removable discontinuities?
2. How can we classify discontinuities based on the behavior of limits?
3. Why is a jump discontinuity non-removable?
4. How do you find removable discontinuities by factoring?
5. Can a function be continuous if it has a removable discontinuity?

**Tip:** Always check both the numerator and the denominator when identifying discontinuities in rational functions!

Which of the following is a function that has a jump discontinuity at x = 2 and a removable discontinuity at x = 4, but is continuous elsewhere?

Learn how to identify functions with a jump discontinuity at x = 2 and a removable discontinuity at x = 4. This problem explores discontinuities in rational functions and how to classify them.

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