Domain of Rational Number

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Full Member
Find domain of H(x) = (3x^2 + x)/(x^2 + 4).

Solution:

I understand the domain of a function defines the values of x that we can put into the function in order to get a valid output.

With rational functions,the textbook explains that we are concerned with the denominator, as the value of this cannot be zero.

Example:

For H(x) = f(x)/g(x), g(x) cannot be zero because division by zero is undefined.

Hence, the values of x NOT in the domain is defined when:

x^2+ 4 = 0

x^2 = - 4

This tells me that there are no real values that satisfy this condition.

"The domain of H(x) is all real values of x with no restriction."

I do not understand the textbook's answer.
There are no real values that satisfy this condition for the function given but at the same time, the domain is all real numbers. Please, explain, in simple terms, the textbook definition....

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pka

Elite Member
Find domain of H(x) = (3x^2 + x)/(x^2 + 4).
The book's answer is "The domain of H(x) is all real values of x with no restriction." CORRECT
I do not understand the textbook's answer.
When dealing with real value functions there are only two restrictions: no division by zero & no even roots of negative numbers.
Well there are no roots involved here. Only $$\displaystyle x^2+4$$ is a divisor and it cannot be zero. So the domain is all real numbers.

Full Member
When dealing with real value functions there are only two restrictions: no division by zero & no even roots of negative numbers.
Well there are no roots involved here. Only $$\displaystyle x^2+4$$ is a divisor and it cannot be zero. So the domain is all real numbers.
Can you explain it another way? You said that only "...x^2 + 4 is a divisor and it cannot be zero." I do not understand why the answer is ALL REAL NUMBERS when the roots of x^2 + 4 are in terms of complex numbers?

Harry_the_cat

Senior Member
Is there any real value of x that makes the denominator equal to zero? The answer is no, so there is no restriction on what x can be, so x can be any real number.

Full Member
Is there any real value of x that makes the denominator equal to zero? The answer is no, so there is no restriction on what x can be, so x can be any real number.
Thank you. Perfect. I now get it.

Otis

Senior Member
... why the [rational function's domain] is ALL REAL NUMBERS when the roots of x^2 + 4 are ... complex numbers?
It seems like you were thinking that roots and domain involve the same numbers. Those are different sets.

Subhotosh Khan

Super Moderator
Staff member
When dealing with real value functions there are only two restrictions: no division by zero & no even roots of negative numbers.
Well there are no roots involved here. Only $$\displaystyle x^2+4$$ is a divisor and it cannot be zero. So the domain is all real numbers.

No logarithm of negative numbers.

No arcsin or arccos of numbers greater than 1 (and corresponding restrictions to secant and cosecant functions)

I think that's all.

topsquark

Full Member
$$\displaystyle e^x \neq 0$$ for any x.

-Dan