Performing Operations on Functions

1. Algebraic Combinations of Functions

Just as we can add, subtract, multiply, and divide numbers, we can perform the same operations on functions. The idea is to create a new function whose output, for any given input $x$, is the result of applying the arithmetic operation to the outputs of the original functions.

The following resource, which is based on your textbook, provides the formal definitions for these operations.

As you saw in the reading, the definitions are quite intuitive:

Given two functions, $f$ and $g$:

• Sum: $(f+g)(x)=f(x)+g(x)$
• Difference: $(f-g)(x)=f(x)-g(x)$
• Product: $(fg)(x)=f(x)g(x)$
• Quotient: $\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}$, provided $g(x)\ne 0$

Let's see these operations in action with a quick video walkthrough that demonstrates the mechanics of combining like terms and applying the FOIL method for multiplication.

2. Graphical Interpretation

Combining functions also has a clear graphical meaning. For addition, the graph of $y=(f+g)(x)$ is created by vertically summing the heights of the individual graphs of $y=f(x)$ and $y=g(x)$ at every value of $x$.

For example, if at $x=2$, we have $f(2)=3$ and $g(2)=-1$, then the new function $(f+g)$ will have the point $(2,2)$ because $3+(-1)=2$. You can see this principle in the image above. Where one of the original functions crosses the x-axis (its value is zero), the sum function is equal to the value of the other function.

3. The Critical Role of Domain

This is the most important—and often trickiest—part of combining functions. When you create a new function, you must determine its domain. A common mistake is to find the final simplified form of the new function and determine its domain from that alone. This is incorrect. The new function inherits the restrictions of its "parent" functions.

Professor Leonard provides an excellent and memorable explanation of this concept. He emphasizes that the domain restrictions of the original functions are like "baggage" or "skeletons in the closet" that you can never get rid of, even if they seem to disappear after simplification.

Let's formalize the rules that Professor Leonard explained.

Rules for Domains of Combined Functions:

Let the domain of $f$ be $D_f$ and the domain of $g$ be $D_g$. The intersection of these domains, $D_f\cap D_g$, is the set of all $x$ that are in both domains.

1. The domains of $f+g$, $f-g$, and $fg$ are the intersection of the domains of $f$ and $g$:
   $D_{f+g}=D_{f-g}=D_{fg}=D_f\cap D_g$

2. The domain of the quotient $f/g$ is the intersection of the domains of $f$ and $g$, with the additional restriction that any $x$ that makes the denominator $g(x)$ equal to zero must be excluded.
   $D_{f/g}=\{x\in D_f\cap D_g\mid g(x)\ne 0\}$

Worked Example

Let $f(x)=\sqrt{x+2}$ and $g(x)=x-1$. Let's find the function $\left(\frac{f}{g}\right)(x)$ and its domain.

Step 1: Find the domain of each original function.

• For $f(x)=\sqrt{x+2}$, the expression under the square root must be non-negative.
  $x+2\ge 0⟹x\ge -2$.
  So, $D_f=[-2,\infty )$.
• For $g(x)=x-1$, this is a linear function with no restrictions.
  So, $D_g=(-\infty ,\infty )$.

Step 2: Find the intersection of the domains.

The intersection $D_f\cap D_g$ includes all numbers that are both greater than or equal to -2 AND are real numbers. This is simply $D_f$.
$D_f\cap D_g=[-2,\infty )$.

Step 3: Find the algebraic form of the new function.

$\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{\sqrt{x+2}}{x-1}$

Step 4: Apply the additional restriction for division.

We must exclude any $x$ from the intersection that makes the denominator $g(x)$ equal to zero.
Set $g(x)=0$:
$x-1=0⟹x=1$.

The value $x=1$ is within our intersection $[-2,\infty )$, so we must explicitly exclude it.

Step 5: State the final domain.

The domain of $\left(\frac{f}{g}\right)(x)$ is all numbers in $[-2,\infty )$ except for 1. In interval notation, this is:
$[-2,1)\cup (1,\infty )$

This systematic process ensures you correctly identify all domain restrictions, both those inherited from the original functions and those newly created by the operation itself.

Conclusion

In this lesson, we have learned how to create new functions by combining existing ones through addition, subtraction, multiplication, and division.

Key Takeaways:

• Functions can be combined arithmetically by performing the operation on their outputs for each input $x$.
• Graphically, this can be visualized as combining the y-values of the graphs at each point, such as adding their vertical heights.
• The domain of a combined function is the intersection of the domains of the original functions.
• For division, you must add a crucial extra restriction: any value of $x$ that makes the denominator zero must be excluded from the domain.

We have now seen two major ways to build complex functions: transformations (modifying a single function) and arithmetic combinations (merging multiple functions). In our next lesson, we will explore a third, powerful method: feeding the output of one function directly into the input of another. This is known as function composition, and it is a fundamental concept that is essential for understanding one of the most important rules in calculus, the Chain Rule.