1. **State the problem:** We have a piecewise function $$f(x) = \begin{cases} \frac{2x^2 - 3x - 2}{x - 2}, & x \neq 2 \\ 7, & x = 2 \end{cases}$$ We need to determine the type of discontinuity at $x=2$. 2. **Recall the definition of continuity:** A function $f$ is continuous at $x=a$ if $$\lim_{x \to a} f(x) = f(a).$$ 3. **Find $f(2)$:** Given directly as $f(2) = 7$. 4. **Find the limit $\lim_{x \to 2} f(x)$:** Since $f(x)$ for $x \neq 2$ is $$\frac{2x^2 - 3x - 2}{x - 2},$$ we simplify the numerator: $$2x^2 - 3x - 2 = (2x + 1)(x - 2)$$ 5. Substitute factorization into the function for $x \neq 2$: $$f(x) = \frac{(2x + 1)(x - 2)}{x - 2}$$ 6. Cancel the common factor $x - 2$: $$f(x) = \frac{(2x + 1)\cancel{(x - 2)}}{\cancel{x - 2}} = 2x + 1, \quad x \neq 2$$ 7. Now find the limit: $$\lim_{x \to 2} f(x) = \lim_{x \to 2} (2x + 1) = 2(2) + 1 = 5$$ 8. **Compare limit and function value:** - $\lim_{x \to 2} f(x) = 5$ - $f(2) = 7$ Since the limit exists but is not equal to the function value, there is a removable discontinuity at $x=2$. **Final answer:** (B) $f$ has a removable discontinuity at $x=2$.