1. **State the problem:** We want to find values of $a$ and $b$ such that the piecewise function $$ f(x) = \begin{cases} \frac{1}{x} & x < -1 \\ ax + b & -1 \leq x \leq \frac{1}{2} \\ \frac{1}{x} & x > \frac{1}{2} \end{cases} $$ is continuous everywhere. 2. **Recall the 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: $\lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} \frac{1}{x} = \frac{1}{-1} = -1$ - Right limit: $\lim_{x \to -1^+} f(x) = a(-1) + b = -a + b$ - Function value: $f(-1) = a(-1) + b = -a + b$ Set left and right limits equal for continuity: $$ -1 = -a + b $$ 4. **Check continuity at $x = \frac{1}{2}$:** - Left limit: $\lim_{x \to (1/2)^-} f(x) = a\left(\frac{1}{2}\right) + b = \frac{a}{2} + b$ - Right limit: $\lim_{x \to (1/2)^+} f(x) = \frac{1}{\frac{1}{2}} = 2$ - Function value: $f\left(\frac{1}{2}\right) = a\left(\frac{1}{2}\right) + b = \frac{a}{2} + b$ Set left and right limits equal for continuity: $$ \frac{a}{2} + b = 2 $$ 5. **Solve the system of equations:** $$ \begin{cases} -a + b = -1 \\ \frac{a}{2} + b = 2 \end{cases} $$ From the first equation: $$ b = a - 1 $$ Substitute into the second: $$ \frac{a}{2} + (a - 1) = 2 $$ Simplify: $$ \frac{a}{2} + a - 1 = 2 $$ $$ \frac{3a}{2} = 3 $$ $$ a = 2 $$ Then, $$ b = 2 - 1 = 1 $$ **Final answer:** $$ a = 2, \quad b = 1 $$