Answer:
Domain: { x ∈ R | x ≠ -2, 3 }
Step-by-step explanation:
[tex]f(x)=\dfrac{(x+2)(x-1)}{(x-3)(x+2)}[/tex]
The domain (x-values) of a rational function is all real numbers of x with the exception of those for which the denominator is 0. Therefore, to find the values of x that need to be excluded from the domain, equate the denominator to zero and solve for x.
The denominator for the given function is [tex](x - 3)(x + 2)[/tex]
[tex]\implies (x - 3)(x + 2)=0[/tex]
[tex]\implies (x - 3)=0\implies x=3[/tex]
[tex]\implies (x + 2)=0 \implies x=-2[/tex]
There will be a hole at x = -2 and an asymptote at x = 3.
So the values of x that need to be excluded from the domain are:
x = 3 and x = -2
Therefore, the domain of the given rational function is: { x ∈ R | x ≠ -2, 3 }
