Answer:
The value of x = -1 makes the denominator of the function equal to zero. That is why this value is not included in the domain of f(x)
Step-by-step explanation:
We have the following expression
[tex]f(x) = \frac{x^2+4x+3}{x^2-x-2}[/tex]
Since the division between zero is not defined then the function f(x) can not include the values of x that make the denominator of the function zero.
Now we search that values of x make 0 the denominator factoring the polynomial [tex]x^2-x-2[/tex]
We need two numbers that when adding them get as a result -1 and when multiplying those numbers, obtain -2 as a result.
These numbers are -2 and 1
Then the factors are:
[tex](x-2) (x + 1)[/tex]
We do the same with the numerator
[tex]x^2+4x+3[/tex]
We need two numbers that when adding them get as a result 4 and when multiplying those numbers, obtain 3 as a result.
These numbers are 3 and 1
Then the factors are:
[tex](x+3)(x + 1)[/tex]
Therefore
[tex]f(x) = \frac{(x+3)(x+1)}{(x-2)(x+1)}[/tex]
Note that [tex]\frac{(x+1)}{(x+1)}=1[/tex] only if [tex]x \neq -1[/tex]
So since [tex]x = -1[/tex] is not included in the domain the function has a discontinuity in [tex]x = -1[/tex]
