Answer:
a) [tex]( -[/tex]∞[tex], -\frac{4}{3}[/tex][tex])[/tex] ∪ [tex](-\frac{4}{3}[/tex] , ∞[tex])[/tex]
b) [tex][-\frac{5}{2} ,[/tex] ∞)
Step-by-step explanation:
The domain of a function is known as the input. We need to find whatever set of numbers can be plugged into every x of a function so that the result is a real number.
a) This problem is a fraction alone, so we need to find out what input of x will make the denominator zero - that will be the number we "skip" when writing out the domain. This is because if the denominator were zero, then it would be 1 divided by 0, which would be undefined and therefore not a real number.
To find this number, we can just take the denominator and solve for x:
[tex]3x + 4 = 0[/tex]
[tex]3x = -4\\x = -\frac{4}{3}[/tex]
Therefore [tex](-[/tex]∞, [tex]-\frac{4}{3}[/tex]) ∪ [tex](-\frac{4}{3},[/tex] ∞[tex])[/tex] is our answer.
b) This problem is just a square root alone, so we need to find out what input of x will make the number inside the square root negative - that number, and the numbers less than it, will be the numbers that we "skip." This is because if the number inside the square root was negative, then it would produce an imaginary answer and not a real one.
A simple trick to find the domain would be to set the terms inside the square root greater than or equal to 0 then solve for x:
[tex]2x + 5\geq 0 \\2x \geq -5\\\\x \geq \frac{-5}{2}[/tex]
Thus, the domain would be any number greater than or equal to [tex]-\frac{5}{2}[/tex], which in interval notation would be written [tex][-\frac{5}{2} ,[/tex] ∞).
