Question

Does anyone know how to do this?? Help please!!!!

Answer 1

When we have a rational function like:

`r(x) = (x + 1)/(x^2 + 3)`

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:

`(1)/(4) = (x + 1)/(x^2 + 3)`

We can multiply both sides by (x^2 + 3)

`(1)/(4)*(x^2 + 3) = (x + 1)/(x^2 + 3)*(x^2 + 3)`

`(x^2 + 3)/(4) = x + 1`

Now we can multiply both sides by 4:

`(x^2 + 3)/(4)*4 = (x + 1)*4`

`x^2 + 3 = 4*x + 4`

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:

`x = (-b +- √(b^2 - 4*a*c) )/(2*a)`

here we have:

a = 1

b = -4

c = -1

Then in this case the solutions are:

`x = (-(-4) +- √((-4)^2 - 4*1*(-1)) )/(2*(1)) = (4 +- 4.47)/(2)`

x = (4 + 4.47)/2 = 4.235

x = (4 - 4.47)/2 = -0.235