the function f is defined above. which of the following statements is true? a f is continuous at x = 2. b f…

Explanation:

Step1: Simplify the function for $x\neq2$

Factor the numerator of $\frac{2x^{2}-3x - 2}{x - 2}$. We have $2x^{2}-3x - 2=(2x + 1)(x - 2)$. So $\frac{2x^{2}-3x - 2}{x - 2}=\frac{(2x + 1)(x - 2)}{x - 2}=2x + 1$ for $x\neq2$.

Step2: Find the limit as $x$ approaches 2

$\lim_{x\rightarrow2}\frac{2x^{2}-3x - 2}{x - 2}=\lim_{x\rightarrow2}(2x + 1)$. Substitute $x = 2$ into $2x+1$, we get $\lim_{x\rightarrow2}(2x + 1)=2\times2+1 = 5$.

Step3: Evaluate the function at $x = 2$

$f(2)=7$.

Step4: Analyze the continuity

Since $\lim_{x\rightarrow2}f(x)=5$ and $f(2)=7$, $\lim_{x\rightarrow2}f(x)\neq f(2)$. But the limit exists. A removable - discontinuity occurs when the limit of a function exists at a point but the function is either undefined or has a different value at that point.

Answer:

B. $f$ has a removable discontinuity at $x = 2$