Answer:
To determine the open intervals on which the function \(f(x) = \frac{26}{x^2 + 12}\) is defined, we need to find the values of \(x\) for which the denominator \(x^2 + 12\) is not equal to zero, because division by zero is undefined.
So, we need to solve the equation \(x^2 + 12 = 0\):
\[x^2 + 12 = 0\]
Subtract 12 from both sides:
\[x^2 = -12\]
Now, take the square root of both sides. Remember that when you take the square root of a negative number, you introduce imaginary numbers:
\[x = \pm \sqrt{-12}\]
The square root of a negative number is represented as \(i\sqrt{-a}\), where \(i\) is the imaginary unit and \(\sqrt{-a}\) is the square root of the positive value of \(-a\). In this case, \(\sqrt{-12} = i\sqrt{12} = 2i\sqrt{3}\).
So, we have two complex solutions:
\[x = 2i\sqrt{3}\]
\[x = -2i\sqrt{3}\]
These are the values of \(x\) for which the denominator \(x^2 + 12\) is equal to zero. Since these values are complex and the function \(f(x)\) is defined for all real numbers except where the denominator is zero, the open intervals on which \(f(x)\) is defined are all real numbers except for the points \(x = 2i\sqrt{3}\) and \(x = -2i\sqrt{3}\).
In interval notation, the open intervals on which \(f(x)\) is defined are:
\((-\infty, -2i\sqrt{3})\) and \((2i\sqrt{3}, \infty)\)
Explanation: