https://file.aiquickdraw.com/mathbot/file/1727119856863c4itxkl9.jpg

Let's analyze the piecewise function \( f(x) \) and the given questions based on the image:

The function is defined as:
\[
f(x) = 
\begin{cases}
-x + b, & \text{if } x < 1 \\
5, & \text{if } x = 1 \\
\frac{-20}{x - b} + 1, & \text{if } x > 1 \text{ (and } x \neq b\text{)}
\end{cases}
\]

### (a) For what value(s) of \( b \) is \( f \) continuous at 1?

Given that \( b = 6 \), this value ensures continuity at \( x = 1 \). This is correct since all parts of the function align when \( b = 6 \), satisfying the continuity condition at \( x = 1 \).

---

### (b) For what value(s) of \( b \) does \( f \) have a removable discontinuity at 1?

A removable discontinuity happens when the limit exists but does not match the value of the function. To check for this:

1. **Left-hand limit**: 
   \[
   \lim_{x \to 1^-} (-x + b) = -1 + b
   \]
   
2. **Right-hand limit**: 
   \[
   \lim_{x \to 1^+} \left( \frac{-20}{x - b} + 1 \right)
   \]

For a removable discontinuity, the limits must be equal but not necessarily equal to the function value \( f(1) = 5 \). We need to solve:
\[
-1 + b = 5 \quad \Rightarrow \quad b = 6
\]
Thus, there is no removable discontinuity at \( x = 1 \) when \( b = 6 \), as this value makes the function continuous.

So, **no other values of \( b \) lead to a removable discontinuity**, as continuity is already ensured for \( b = 6 \).

---

### (c) For what value(s) of \( b \) does \( f \) have an infinite discontinuity at 1?

An infinite discontinuity at \( x = 1 \) would occur if the function blows up to infinity as \( x \to 1^+ \). For that, we need the denominator \( x - b \) to approach zero as \( x \to 1 \). 

This would occur if \( b = 1 \), since:

\[
\lim_{x \to 1^+} \frac{-20}{x - 1} = -\infty
\]

So, the value of \( b \) that creates an infinite discontinuity at \( x = 1 \) is \( b = 1 \).

---

### (d) For what value(s) of \( b \) does \( f \) have a jump discontinuity at 1?

A jump discontinuity occurs if the left-hand and right-hand limits exist but are not equal. We already computed:

- Left-hand limit: \( \lim_{x \to 1^-} (-1 + b) = b - 1 \)
- Right-hand limit: \( \lim_{x \to 1^+} \frac{-20}{x - b} + 1 \)

We solve for when these two limits are not equal and \( f(x) \) jumps. To find the intervals for which the function has a jump discontinuity, we use a **process of elimination** based on the other cases (continuity, removable, and infinite discontinuities).

Thus, **\( b \neq 6 \)** and \( b \neq 1 \) could lead to a jump discontinuity.

---

Would you like more details on any of the above solutions or need help with another question?

### Related Questions:
1. Why does \( b = 6 \) ensure continuity at \( x = 1 \)?
2. What is the difference between removable and jump discontinuities?
3. How do limits help identify the type of discontinuity?
4. Could there be any other values of \( b \) where the function is discontinuous?
5. How would you check if the function is differentiable at \( x = 1 \)?

### Tip:
Always check both the left-hand and right-hand limits carefully to classify discontinuities at a point.

For what value(s) of b does the function f have continuity, removable discontinuity, infinite discontinuity, or jump discontinuity at x = 1?

Explore the analysis of a piecewise function to determine the values of b for which it is continuous, has a removable discontinuity, infinite discontinuity, or jump discontinuity at x = 1. Suitable for students in advanced high school or introductory college calculus.

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