Question

Find the domain of 1/(x²-1)(x+3)

Answer 1

Answer:

Domain of $\dfrac{1}{(x^2-1)(x+3)}$ is the set $\mathbb{R}\setminus \{1,-1,-3\}$

Step-by-step explanation:

The domain of a function is the set where it is defined.

Given the function

$\dfrac{1}{(x^2-1)(x+3)}$

This function is not defined when the denominator is zero. That is, it is not defined when,

$(x^2-1)(x+3) =0$

On solving the above equation, we get,

$(x^2-1) =0 \ \ \ \ or \ \ \ \ x+3=0\\\implies x^2=1 \ \ \ \ or \ \ \ x=0-3\\\implies x=1, -1 \ \ \ \ or \ \ \ \ x=-3$

So the function is not defined when $x=1,-1,-3$.

Thus the domain is set of all real numbers except the numbers 1, -1 and -3.

Hence the domain is $\mathbb{R}\setminus \{1,-1,-3\}$.