Given a function
`f(x){{:(-1,if,xle0),(ax+b,if,0ltxlt1),(1,if,xge1):}`
where a, b are constant. The function is continuous everywhere.
What is the value of b?

Options

A. `-1`
B. 1
C. 0
D. 2

Answer

To find the value of \( b \) in the given piecewise function

\[ f(x) = \begin{cases} -1 & \text{if } x \leq 0 \\ ax + b & \text{if } 0 < x < 1 \\ 1 & \text{if } x \geq 1 \end{cases} \]

we need to ensure that the function is continuous everywhere. This means the left-hand limit and right-hand limit at the points where the function changes must be equal to the function's value at those points.

### Step 1: Check continuity at \( x = 0 \)

1. **Left-hand limit as \( x \) approaches 0**:
\[ \lim_{x \to 0^-} f(x) = -1 \]

2. **Right-hand limit as \( x \) approaches 0**:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0} (ax + b) = a(0) + b = b \]

For the function to be continuous at \( x = 0 \), these two limits must be equal:
\[ -1 = b \]

### Step 2: Conclusion

From the above steps, we conclude that the value of \( b \) is:
\[ b = -1 \]

### Final Answer:
The value of \( b \) is \(-1\).

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