91Ó°ÊÓ

A function's domain refers to the set of all possible input values (x-values) for which the function is defined. Understanding this concept is crucial because plugging in an x-value outside of the domain will result in an undefined expression.
For the functions provided in the exercise, we have:
1. Function f(x): Given by: \(f(x) = 1 + \frac{1}{x}\). Here, division by zero is undefined, so x cannot be 0. Thus, the domain of f(x) is all real numbers except x=0.
2. Function g(x): Given by: \(g(x) = \frac{1}{x}\). Similarly, for g(x), division by zero is also undefined, so x cannot be 0. The domain of g(x) is all real numbers except x=0.

When performing operations on functions, always consider the intersection of their domains. This means finding values of x for which both functions are defined.

When adding or subtracting functions, we combine them as follows:

• Addition: \((f+g)(x) = f(x) + g(x)\)
  For example, with our given functions:
  \((f+g)(x) = 1 + \frac{1}{x} + \frac{1}{x} = 1+ \frac{2}{x}\).
  The domain remains the same i.e., all real numbers except x=0.


• Subtraction: \((f-g)(x) = f(x) - g(x)\)
  For our example:
  \((f-g)(x) = 1 + \frac{1}{x} - \frac{1}{x} = 1\).
  The domain is again, all real numbers except x=0.

Even though the simplified expression may look different, the domain is affected by the original functions.

Multiplying or dividing functions involves creating new expressions:

• Multiplication: \((f \cdot g)(x) = f(x) \times g(x)\)
  In our case:
  \((f \cdot g)(x) = (1 + \frac{1}{x}) \times \frac{1}{x} = \frac{1}{x} + \frac{1}{x^2}\).
  The domain here is the same: all real numbers except x=0.


• Division: \(\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}\)
  For our functions:
  \(\left(\frac{f}{g}\right)(x) = \frac{1 + \frac{1}{x}}{\frac{1}{x}} = x + 1\).
  Here, too, the domain is all real numbers except x=0.

While performing these operations, always ensure that division by zero is avoided by checking the domain.