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 the function $f(x)$ to be continuous at $x = -1$, we must have \[ \lim_{x \to -1} f(x) = f(-1). \] We need to calculate the limit of $f(x)$ as $x$ approaches $-1$. First, simplify the expression for $f(x)$ when $x \neq -1$: \[ f(x) = \frac{x^2 - 2x - 3}{x + 1}. \] Factor the numerator: \[ x^2 - 2x - 3 = (x - 3)(x + 1). \] Thus, \[ f(x) = \frac{(x - 3)(x + 1)}{x + 1}. \] For $x \neq -1$, the $(x + 1)$ terms cancel out, and we are left with: \[ f(x) = x - 3. \] Now, compute the limit as $x$ approaches $-1$: \[ \lim_{x \to -1} f(x) = \lim_{x \to -1} (x - 3) = -1 - 3 = -4. \] For continuity at $x = -1$, we need $f(-1) = k$. Hence, \[ k = -4. \]