91Ó°ÊÓ

When we talk about adding or subtracting functions, it involves creating a new function by adding or subtracting the outputs of two given functions for each input value. Suppose we have two functions, like in our exercise, where \( f(x) = \frac{x}{x+1} \) and \( g(x) = x^3 \). Function addition results in \((f+g)(x) = f(x) + g(x) = \frac{x}{x+1} + x^3\). This is like combining the effects of two separate operations into one.For subtraction, we follow a similar approach. We subtract the output of one function from the output of the other. This yields \((f-g)(x) = f(x) - g(x) = \frac{x}{x+1} - x^3\). In both addition and subtraction, the result is a new function. Important to remember is that the operations are performed separately on the corresponding outputs of the functions for each input value.

• Combine functions by adding or subtracting outputs for each input.
• The domain of the resulting function is the intersection of the domains of the original functions.

Multiplying and dividing functions transforms them in different ways compared to addition and subtraction. By multiplying functions, we create a new function by multiplying their outputs for each input value, as in \((fg)(x) = f(x) \cdot g(x) = \frac{x}{x+1} \cdot x^3 = \frac{x^4}{x+1}\). Here, each output of \(f(x)\) is multiplied by the corresponding output of \(g(x)\).Division of functions involves dividing the output of one function by the output of the other function, leading to another new function. In our example: \((f/g)(x) = \frac{f(x)}{g(x)} = \frac{x}{x^3(x+1)} = \frac{1}{x^2(x+1)}\). It’s crucial to be mindful of division because it affects the domain and entails restrictions where the denominator cannot be zero.

• Multiply by multiplying outputs for each input.
• Divide by dividing outputs, ensuring the denominator isn't zero.

Understanding a function's domain is essential, as it defines all the possible input values for which the function is defined. For both individual functions and operations like addition, subtraction, multiplication, and division, determining the domain involves identifying restrictions on the input values.In our division example, where \((f/g)(x) = \frac{1}{x^2(x+1)}\), the domain excludes any values that make the denominator zero, leading to \(x^2(x+1) = 0\). Solving this, we find \(x = 0\) and \(x = -1\) are problematic values because they cause division by zero, which is undefined. Thus, the domain of \((f/g)(x)\) is all real numbers excluding these values.

• Domain refers to the allowable inputs of a function.
• Exclude inputs that result in undefined operations, such as dividing by zero.
• Check all parts of the function to identify restrictions.