Question

Explain why f(x) = x^2+4x+3/x^2-x-2 is not continuous at x = -1.

Answer 1

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

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

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 $x^2-x-2$

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:

$(x-2) (x + 1)$

We do the same with the numerator

$x^2+4x+3$

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:

$(x+3)(x + 1)$

Therefore

$f(x) = \frac{(x+3)(x+1)}{(x-2)(x+1)}$

Note that $\frac{(x+1)}{(x+1)}=1$ only if $x \neq -1$

So since $x = -1$ is not included in the domain the function has a discontinuity in $x = -1$