1. **Problem statement:** Find the values of $a$ and $b$ such that the function $$ f(x) = \begin{cases} -2, & x \leq -1 \\ ax - b, & -1 < x < 1 \\ 3, & x \geq 1 \end{cases} $$ is continuous everywhere. 2. **Continuity condition:** A function is continuous at a point if the left-hand limit, right-hand limit, and the function value at that point are all equal. 3. **Check continuity at $x = -1$:** - Left limit: $\lim_{x \to -1^-} f(x) = -2$ - Right limit: $\lim_{x \to -1^+} f(x) = a(-1) - b = -a - b$ - Function value: $f(-1) = -2$ Set left and right limits equal for continuity: $$ -2 = -a - b $$ 4. **Check continuity at $x = 1$:** - Left limit: $\lim_{x \to 1^-} f(x) = a(1) - b = a - b$ - Right limit: $\lim_{x \to 1^+} f(x) = 3$ - Function value: $f(1) = 3$ Set left and right limits equal for continuity: $$ a - b = 3 $$ 5. **Solve the system of equations:** $$ \begin{cases} -2 = -a - b \\ 3 = a - b \end{cases} $$ Add the two equations: $$ -2 + 3 = (-a - b) + (a - b) \implies 1 = -2b \implies b = -\frac{1}{2} $$ Substitute $b = -\frac{1}{2}$ into $3 = a - b$: $$ 3 = a - \left(-\frac{1}{2}\right) = a + \frac{1}{2} \implies a = 3 - \frac{1}{2} = \frac{5}{2} $$ 6. **Final answer:** $$ a = \frac{5}{2}, \quad b = -\frac{1}{2} $$ These values make $f(x)$ continuous everywhere.