Question

1. let $f(x)=\frac{x^{2}-4x + 3}{x - 3}$. how could we define $f(3)$ so that $f$ is continuous at $x = 3$?

Explanation:

Step1: Factor the numerator

We factor $x^{2}-4x + 3$ as $(x - 3)(x - 1)$. So $f(x)=\frac{(x - 3)(x - 1)}{x - 3}$, for $x
eq3$.

Step2: Simplify the function

Cancel out the common factor $(x - 3)$ in the numerator and denominator. We get $f(x)=x - 1$ for $x
eq3$.

Step3: Find the limit as $x$ approaches 3

We find $\lim_{x
ightarrow3}f(x)$. Since $f(x)=x - 1$ for $x
eq3$, then $\lim_{x
ightarrow3}f(x)=\lim_{x
ightarrow3}(x - 1)$. Substitute $x = 3$ into $x - 1$, we have $\lim_{x
ightarrow3}(x - 1)=3-1 = 2$.

Step4: Define $f(3)$ for continuity

For a function to be continuous at $x=a$, $\lim_{x
ightarrow a}f(x)=f(a)$. So to make $f(x)$ continuous at $x = 3$, we define $f(3)=2$.

Answer:

$f(3)=2$