For this case we have to:
Represents the solution of all strict minor numbers to -5.
Represents the solution of all strict major numbers to 5.
The global solution is given by the intersection of both sets. The symbol of intersection is denoted as: ∩
If we represent the solution in a numerical line, it gives us empty. do not intersect.
Answeer:
Empty set
Answer:
When we have a rational function like:
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:
We can multiply both sides by (x^2 + 3)
Now we can multiply both sides by 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:
here we have:
a = 1
b = -4
c = -1
Then in this case the solutions are:
x = (4 + 4.47)/2 = 4.235
x = (4 - 4.47)/2 = -0.235