Question

Find $k$ so that \[ f(x) = \begin{cases} \frac{x^2 - 2x - 3}{x + 1}, & \text{if } x \neq -1 \\ k, & \text{if } x = -1 \end{cases} \] is continuous at $x = -1$.

Hint:

For a function to be continuous at a point, the limit of the function as $x$ approaches the point must equal the value of the function at that point.

Answer 1

For $f(x)$ to be continuous at $x = -1$, the limit of $f(x)$ as $x$ approaches $-1$ must equal $f(-1)$. We calculate this limit. For $x eq -1$, the function is given by $f(x) = \frac{x^2 - 2x - 3}{x + 1}$. Factoring the numerator yields $x^2 - 2x - 3 = (x - 3)(x + 1)$. Substituting this into the expression for $f(x)$, we get $f(x) = \frac{(x - 3)(x + 1)}{x + 1}$. For $x eq -1$, we can cancel the $(x + 1)$ terms, simplifying $f(x)$ to $f(x) = x - 3$. The limit is then $\lim_{x \to -1} f(x) = \lim_{x \to -1} (x - 3) = -1 - 3 = -4$. For continuity at $x = -1$, $f(-1)$ must equal this limit. Therefore, $f(-1) = k = -4$.