How would you "remove the discontinuity" of f ? In other words, how would you define f(5) in order to make f continuous at 5?f(x) = x2 − 4x − 5x − 5 f(5) =

To make the function $f(x) = \frac{x^2 - 4x - 5}{x - 5}$ continuous at $x = 5$, we need to first identify the limit as $x$ approaches 5 and then set $f(5)$ equal to that limit.

Steps to Remove the Discontinuity

1. Break Down the Problem

   • Identify the type of discontinuity at $x = 5$.
   • Calculate the limit of $f(x)$ as $x$ approaches 5.

2. Relevant Concepts

   • We will use the limit definition to find $\lim_{x \to 5} f(x)$.

3. Analysis and Detail

   • Factor the numerator:

   $x^2 - 4x - 5 = (x - 5)(x + 1)$

   • Then rewrite $f(x)$:

   $f(x) = \frac{(x - 5)(x + 1)}{(x - 5)} \quad \text{for } x \neq 5$

   • Thus, we can simplify $f(x)$:

   $f(x) = x + 1 \quad \text{for } x \neq 5$

   • Now, calculate the limit:

   $\lim_{x \to 5} f(x) = \lim_{x \to 5} (x + 1) = 5 + 1 = 6$

4. Verify and Summarize

   • To remove the discontinuity, we need to set $f(5)$ equal to the limit:

   $f(5) = 6$

Final Answer

To make $f$ continuous at $x = 5$, define $f(5) = 6$.