Question

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

Hint:

For piecewise functions to be continuous at a certain point, the function values from both sides of the point must match the function value at that point. Simplify the expression and find the limit to determine the necessary value.

Answer 1

For continuity at $ x = -1 $, the limit of the function as $ x $ approaches $ -1 $ must equal the function's value at $ x = -1 $. Mathematically, this is expressed as:\[\lim_{x \to -1} f(x) = f(-1)\]Step 1: Simplify the expression for $ f(x) $ where $ x eq -1 $:\[f(x) = \frac{x^2 - 2x - 3}{x + 1}\]Factor the numerator:\[x^2 - 2x - 3 = (x - 3)(x + 1)\]Substitute the factored numerator back into the expression for $ f(x) $:\[f(x) = \frac{(x - 3)(x + 1)}{x + 1}\]For $ x eq -1 $, the term $ (x + 1) $ cancels out:\[f(x) = x - 3\]Step 2: Compute the limit as $ x $ approaches $ -1 $:\[\lim_{x \to -1} f(x) = \lim_{x \to -1} (x - 3) = -1 - 3 = -4\]Step 3: For continuity at $ x = -1 $, the function value $ f(-1) $ must equal the limit:\[f(-1) = k = -4\]Therefore, the value of $ k $ is $ \boxed{-4} $.