MHB Finding the Domain of a Function

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To find the domain of the function \(\sqrt{(x^2 + 4)/(x^2 - 4)}\), it is essential to ensure that the expression under the square root is non-negative and that the denominator is not zero. The square root requires that \((x^2 + 4)/(x^2 - 4) \geq 0\), while the fraction mandates that \(x^2 - 4 \neq 0\). The denominator is zero at \(x = 2\) and \(x = -2\), which must be excluded from the domain. Additionally, since \(x^2 + 4\) is always positive, the critical consideration is the sign of \(x^2 - 4\). Therefore, the domain excludes \(x = 2\) and \(x = -2\) and includes all other real numbers where the fraction is non-negative.
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How do you find the Domain of \sqrt{(x^2 + 4)/(x^2 - 4)} ?
 
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megacat8921 said:
How do you find the Domain of \sqrt{(x^2 + 4)/(x^2 - 4)} ?

What restrictions are implied by a square root? What restrictions are implied by a fraction?
 
Prove It said:
What restrictions are implied by a square root? What restrictions are implied by a fraction?

There cannot be a negative number under the square root and a fraction cannot have zero as a denominator. Knowing that, I still cannot figure out what steps to take to figure the problem out.
 
megacat8921 said:
There cannot be a negative number under the square root and a fraction cannot have zero as a denominator. Knowing that, I still cannot figure out what steps to take to figure the problem out.

That's EXACTLY what you need!

What values of x will give a zero denominator?

What values of x will give something negative under the square root?
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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