Domain Equals All Real Numbers

[harpazo]

Given f(x) = 5/(x^2 + 1), the domain is all real numbers but I do not know the reason why that is the case.

For rational functions, we set the denominator equal to 0 and solve for the variable. If we set the given denominator to be 0, the following happens:

x^2 + 1 = 0

x^2 = - 1

Stuck here....

 

[Harry_the_cat]

The domain is the set of numbers that x can be. (You might be thinking of the range ie the values that y or f(x) can be.)

Is there any restriction on what x can be? No, so the domain is all real numbers.

 

[Dr.Peterson]

Given f(x) = 5/(x^2 + 1), the domain is all real numbers but I do not know the reason why that is the case.

For rational functions, we set the denominator equal to 0 and solve for the variable. If we set the given denominator to be 0, the following happens:

x^2 + 1 = 0

x^2 = - 1

Stuck here....

This is essentially the same issue as in the problem with [MATH]\sqrt{x^2 + 1}[/MATH] in the denominator.

You are solving [MATH]x^2 = - 1[/MATH]. You know that the square of any real number is non-negative; so there are no real solutions to this equation, therefore no numbers are excluded from the domain (as x is restricted to real numbers in your context).

 

[Otis]

Given f(x) = 5/(x^2 + 1), the domain is all real numbers but I do not know the reason why …

The only denominator that would cause function f to become undefined is zero.

But, there is no Real value of x that will make the denominator zero because x^2+1 is always a positive number.

Graph the parabola y=x^2+1. That graph is the same as y=x^2 shifted vertically upward one unit. You can see, therefore, that the graph of x^2+1 will never touch the x-axis (i.e., it will not equal zero, at any value of x).

?

 

[harpazo]

Thank you everyone. Exactly what I thought.

 

[pka]

Given f(x) = 5/(x^2 + 1), the domain is all real numbers but I do not know the reason why that is the case.

To harpazo. Do you have have any idea whatsoever what this question is all about?
The domain of any function is the set of numbers that make the function defined.
Now there are two basic no-nos: there cannot be division by zero & there cannot be an even root of a negative number.
Look at the function \(\displaystyle \large f(x)=\frac{5}{x^2+1}\).
Does that function violate either of those conditions?
If not, what is your problem harpazo?

 

[harpazo]

To harpazo. Do you have have any idea whatsoever what this question is all about?
The domain of any function is the set of numbers that make the function defined.
Now there are two basic no-nos: there cannot be division by zero & there cannot be an even root of a negative number.
Look at the function \(\displaystyle \large f(x)=\frac{5}{x^2+1}\).
Does that function violate either of those conditions?
If not, what is your problem harpazo?

If we set x^2 + 1 = 0, this happens:

x^2 = -1

sqrt{x^2} = sqrt{-1}

x = i and x = - i

It violates one of the rules you listed above.
Yes, I do understand the question and the domain situation for rational functions.

 

[Harry_the_cat]

x is the variable usually used to indicate real numbers. f(x) is a function where the inputs come from the set of real numbers, or in this case the inputs are the whole set of real numbers.

You are correct to say that x²=1 leads to x = i or x = -i. But here we need to assume that the domain comes from the set of real numbers only.

z is the variable used to indicate complex numbers.

 

[harpazo]

x is the variable usually used to indicate real numbers. f(x) is a function where the inputs come from the set of real numbers, or in this case the inputs are the whole set of real numbers.

You are correct to say that x²=1 leads to x = i or x = -i. But here we need to assume that the domain comes from the set of real numbers only.

z is the variable used to indicate complex numbers.

Absolutely. This is a good review as I continue with Sullivan's textbook.