Answer:
When we have a rational function like:
[tex]r(x) = \frac{x + 1}{x^2 + 3}[/tex]
The domain will be the set of all real numbers, such that the denominator is different than zero.
So the first step is to find the values of x such that the denominator (x^2 + 3) is equal to zero.
Then we need to solve:
x^2 + 3 = 0
x^2 = -3
x = √(-3)
This is the square root of a negative number, then this is a complex number.
This means that there is no real number such that x^2 + 3 is equal to zero, then if x can only be a real number, we will never have the denominator equal to zero, so the domain will be the set of all real numbers.
D: x ∈ R.
b) we want to find two different numbers x such that:
r(x) = 1/4
Then we need to solve:
[tex]\frac{1}{4} = \frac{x + 1}{x^2 + 3}[/tex]
We can multiply both sides by (x^2 + 3)
[tex]\frac{1}{4}*(x^2 + 3) = \frac{x + 1}{x^2 + 3}*(x^2 + 3)[/tex]
[tex]\frac{x^2 + 3}{4} = x + 1[/tex]
Now we can multiply both sides by 4:
[tex]\frac{x^2 + 3}{4}*4 = (x + 1)*4[/tex]
[tex]x^2 + 3 = 4*x + 4[/tex]
Now we only need to solve the quadratic equation:
x^2 + 3 - 4*x - 4 = 0
x^2 - 4*x - 1 = 0
We can use the Bhaskara's formula to solve this, remember that for an equation like:
a*x^2 + b*x + c = 0
the solutions are:
[tex]x = \frac{-b +- \sqrt{b^2 - 4*a*c} }{2*a}[/tex]
here we have:
a = 1
b = -4
c = -1
Then in this case the solutions are:
[tex]x = \frac{-(-4) +- \sqrt{(-4)^2 - 4*1*(-1)} }{2*(1)} = \frac{4 +- 4.47}{2}[/tex]
x = (4 + 4.47)/2 = 4.235
x = (4 - 4.47)/2 = -0.235
