Question

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

Answer 1

Answer:

When we have a rational function like:

$r(x) = \frac{x + 1}{x^2 + 3}$

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:

$\frac{1}{4} = \frac{x + 1}{x^2 + 3}$

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

$\frac{1}{4}*(x^2 + 3) = \frac{x + 1}{x^2 + 3}*(x^2 + 3)$

$\frac{x^2 + 3}{4} = x + 1$

Now we can multiply both sides by 4:

$\frac{x^2 + 3}{4}*4 = (x + 1)*4$

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

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:

$x = \frac{-b +- \sqrt{b^2 - 4*a*c} }{2*a}$

here we have:

a = 1

b = -4

c = -1

Then in this case the solutions are:

$x = \frac{-(-4) +- \sqrt{(-4)^2 - 4*1*(-1)} }{2*(1)} = \frac{4 +- 4.47}{2}$

x = (4 + 4.47)/2 = 4.235

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