Applied Calculus MA25C01 | ✅ Continuity Problem 3 : \[ f(x)=\begin{cases} -2 & \text{if } x \leq -1 \\ ax-b & \text{if } -1 < x < 1 \\ 3 & \text{if } x \geq 1 \end{cases} \]

▶ Tutorial Video:Continuity -Example 3

Finding Constants for Function Continuity

Reference: [A/M-22-R-21]

Example: For what values of \(a\) and \(b\), is the function \(f(x)\) continuous at every \(x\)?

\[ f(x)=\begin{cases} -2 & x \leq -1 \\ ax-b & -1 < x < 1 \\ 3 & x \geq 1 \end{cases} \]

Solution

For a function to be continuous at a point \(x = c\), the following condition must be met:

\[ \lim_{x \to c^{-}} f(x) = \lim_{x \to c^{+}} f(x) = f(c) \]

---

Step 1: Check Continuity at \(x = -1\)

• Left-hand limit: \( \lim_{x \to -1^{-}} f(x) = -2 \)
• Right-hand limit: \( \lim_{x \to -1^{+}} f(x) = a(-1) - b = -a - b \)

Setting them equal for continuity:

\( -a - b = -2 \implies \mathbf{a + b = 2} \) — (Eq. 1)

---

Step 2: Check Continuity at \(x = 1\)

• Left-hand limit: \( \lim_{x \to 1^{-}} f(x) = a(1) - b = a - b \)
• Right-hand limit: \( \lim_{x \to 1^{+}} f(x) = 3 \)

Setting them equal for continuity:

\( \mathbf{a - b = 3} \) — (Eq. 2)

---

Step 3: Solving the System of Equations

We now solve the simultaneous equations:

1. \( a + b = 2 \)
2. \( a - b = 3 \)

Adding the two equations:

\[ (a + b) + (a - b) = 2 + 3 \]

\[ 2a = 5 \implies \mathbf{a = \frac{5}{2}} \]

Substituting \(a\) back into Eq. 1:

\[ \frac{5}{2} + b = 2 \implies b = 2 - \frac{5}{2} \]

\[ \mathbf{b = -\frac{1}{2}} \]

Final Answer:
\[ a = \frac{5}{2}, \quad b = -\frac{1}{2} \]

✔ —————————————————————— ✔