Question

for the function f, how would you remove the discontinuity? in other words, how would you define f(4) in order for f to be continuous at 4

f(x) = \frac{x^2 - 3x - 4}{x - 4}

f(4) = \boxed{\space}

Explanation:

Step1: Factor the numerator

To simplify the function \( f(x)=\frac{x^{2}-3x - 4}{x - 4} \), we factor the quadratic expression in the numerator. We need two numbers that multiply to \(-4\) and add to \(-3\). Those numbers are \(-4\) and \(1\). So, \(x^{2}-3x - 4=(x - 4)(x+1)\).
So the function becomes \( f(x)=\frac{(x - 4)(x + 1)}{x - 4} \) (for \(x
eq4\)).

Step2: Cancel the common factor

We can cancel out the common factor of \(x - 4\) in the numerator and the denominator (since \(x
eq4\), we are not dividing by zero here). After canceling, we get \( f(x)=x + 1 \) for \(x
eq4\).

Step3: Find the limit as \(x\) approaches 4

To make the function continuous at \(x = 4\), we need to define \(f(4)\) as the limit of \(f(x)\) as \(x\) approaches \(4\). We evaluate the limit of \(x + 1\) as \(x\) approaches \(4\). Using the direct substitution property (since \(x+1\) is a polynomial), we substitute \(x = 4\) into \(x + 1\).
\(\lim_{x
ightarrow4}(x + 1)=4+1 = 5\).

Answer:

\(5\)