Answer:
To locate the discontinuities of the function \(y = \frac{1}{3 + e^{2/x}}\), we need to find the values of \(x\) for which the function is not defined or where it approaches infinity.
The potential discontinuities occur when the denominator of the function equals zero because division by zero is undefined. So, we need to solve for \(x\) in the equation \(3 + e^{2/x} = 0\).
First, let's isolate \(e^{2/x}\) on one side:
\[e^{2/x} = -3\]
Now, take the natural logarithm (ln) of both sides:
\[\ln(e^{2/x}) = \ln(-3)\]
Using the property that \(\ln(e^a) = a\), we get:
\[\frac{2}{x} = \ln(-3)\]
Now, isolate \(x\):
\[x = \frac{2}{\ln(-3)}\]
However, this value is not a real number because the natural logarithm of a negative number is undefined for real numbers. Therefore, the function \(y = \frac{1}{3 + e^{2/x}}\) has a discontinuity at \(x = \frac{2}{\ln(-3)}\), and this point is not included in the domain of the function for real numbers. The function is defined for all other real values of \(x\).
Explanation: