Functions and Their Graphs
Deepen your understanding of functions as the foundation for everything ahead
You have been using functions your entire life. When you check the weather forecast and decide what to wear, you are taking an input (temperature) and producing an output (clothing choice). When you calculate a tip at a restaurant, you are applying a function to the bill total. Every time you convert between Celsius and Fahrenheit, you are working with a function and its inverse. In Algebra 2, we take this intuition you already have and develop it into powerful mathematical machinery. By the end of this lesson, you will see functions not as abstract rules to memorize, but as tools for describing how things in the world depend on one another. And once you understand functions deeply, everything else in Algebra 2 will make more sense.
Core Concepts
Functions Review: What Makes a Function a Function
Let us start with what you already know. A function is a rule that takes each input and produces exactly one output. You can think of it as a very reliable machine: put in a value, get out a value, and that value is always the same for the same input.
The notation $f(x)$ says: “the function named $f$, when given the input $x$, produces this output.” When you see $f(3) = 10$, it means that when you put 3 into the function $f$, you get 10 out. Every single time.
Domain is the set of all inputs that work, meaning all values you are allowed to feed into the function. Range is the set of all outputs that can come out.
Why does this matter? Because functions give us predictability. If you know a function and you know an input, you know exactly what the output will be. That reliability is what makes functions so useful for modeling the real world.
Analyzing Functions: Three Windows into the Same Idea
You can understand a function through three different lenses: its equation, its graph, and a table of values. Each view reveals something different.
From an equation like $f(x) = x^2 - 4$, you can:
- Calculate any output you want by plugging in values
- See the algebraic structure (this is a polynomial, specifically a quadratic)
- Find domain restrictions (what values make the equation undefined?)
From a graph, you can:
- See the overall shape and behavior at a glance
- Identify where the function increases, decreases, or stays flat
- Spot maximum and minimum values
- Observe symmetry patterns
From a table, you can:
- See specific input-output pairs
- Notice patterns in how outputs change as inputs change
- Estimate values between the ones shown
The skill of moving fluently between these three representations is one of the most important abilities you will develop in Algebra 2.
Increasing, Decreasing, and Constant Intervals
When you look at a function’s graph from left to right (as $x$ increases), the output might be going up, going down, or staying flat. We have precise language for this:
Increasing: The function is increasing on an interval if, as $x$ gets larger, $f(x)$ also gets larger. On the graph, the curve goes uphill as you move right.
Decreasing: The function is decreasing on an interval if, as $x$ gets larger, $f(x)$ gets smaller. On the graph, the curve goes downhill as you move right.
Constant: The function is constant on an interval if $f(x)$ stays the same regardless of $x$. On the graph, this appears as a horizontal segment.
Here is the key: we describe these behaviors using intervals of $x$-values, not $y$-values. If a function increases from $x = -2$ to $x = 3$, we write “increasing on $(-2, 3)$.”
For example, the function $f(x) = x^2$ is:
- Decreasing on $(-\infty, 0)$ (going downhill as $x$ moves toward 0 from the left)
- Increasing on $(0, \infty)$ (going uphill as $x$ moves away from 0 to the right)
Maximum and Minimum Values
Functions often have high points and low points. We distinguish between two types:
Local (or relative) maximum: A point where the function value is higher than all nearby points. Think of it as the top of a hill. There might be higher hills elsewhere, but this is the highest spot in the immediate neighborhood.
Local (or relative) minimum: A point where the function value is lower than all nearby points. This is the bottom of a valley.
Absolute (or global) maximum: The highest point on the entire function. No other point anywhere has a higher output.
Absolute (or global) minimum: The lowest point on the entire function. No other point anywhere has a lower output.
A function can have many local maxima and minima, but at most one absolute maximum and one absolute minimum (though it might have neither if the function extends to infinity).
For example, $f(x) = x^2$ has:
- An absolute minimum at $(0, 0)$, meaning the smallest output is 0, which occurs when $x = 0$
- No absolute maximum (the function grows forever as $x$ moves away from 0)
- No local maxima
Even and Odd Functions: Symmetry You Can Exploit
Some functions have beautiful symmetry that makes them easier to understand and work with.
Even functions are symmetric about the $y$-axis. If you fold the graph along the $y$-axis, the two halves match perfectly. Algebraically, a function is even when $f(-x) = f(x)$ for all $x$ in the domain.
The simplest example is $f(x) = x^2$. Check: $f(-3) = (-3)^2 = 9$ and $f(3) = 3^2 = 9$. The outputs are identical for $x$ and $-x$.
Other even functions: $f(x) = x^4$, $f(x) = |x|$, $f(x) = \cos(x)$.
Odd functions are symmetric about the origin. If you rotate the graph 180 degrees around the origin, it looks exactly the same. Algebraically, a function is odd when $f(-x) = -f(x)$ for all $x$ in the domain.
The simplest example is $f(x) = x^3$. Check: $f(-2) = (-2)^3 = -8$ and $f(2) = 2^3 = 8$. The outputs are opposites.
Other odd functions: $f(x) = x$, $f(x) = x^5$, $f(x) = \sin(x)$.
Most functions are neither even nor odd. For example, $f(x) = x^2 + x$ is neither. Check: $f(-1) = 1 - 1 = 0$ and $f(1) = 1 + 1 = 2$. The outputs are not equal (not even) and not opposites (not odd).
Piecewise Functions: Different Rules for Different Inputs
Sometimes a single formula cannot capture a function’s behavior. Real-world situations often have different rules that apply in different circumstances. Enter piecewise functions.
A piecewise function uses different formulas depending on which piece of the domain the input falls into. Here is an example:
$$f(x) = \begin{cases} x^2 & \text{if } x < 0 \ 2x + 1 & \text{if } x \geq 0 \end{cases}$$
This function says: “If the input is negative, square it. If the input is zero or positive, double it and add 1.”
To evaluate $f(-3)$: Since $-3 < 0$, use the first rule: $f(-3) = (-3)^2 = 9$.
To evaluate $f(4)$: Since $4 \geq 0$, use the second rule: $f(4) = 2(4) + 1 = 9$.
To evaluate $f(0)$: Since $0 \geq 0$, use the second rule: $f(0) = 2(0) + 1 = 1$.
Piecewise functions show up constantly in real life:
- Tax brackets (different rates for different income levels)
- Shipping costs (different rates for different weight ranges)
- Cell phone plans (different rates before and after using your data allotment)
- Overtime pay (regular rate up to 40 hours, then time-and-a-half)
Function Composition: Functions Inside Functions
Here is a powerful idea: you can use the output of one function as the input to another. This is called composition.
If you have two functions $f$ and $g$, the composition $(f \circ g)(x)$ means: first apply $g$ to $x$, then apply $f$ to that result.
In symbols: $(f \circ g)(x) = f(g(x))$
Read this as “f of g of x” and evaluate from the inside out.
Example: Let $f(x) = x^2$ and $g(x) = x + 3$.
To find $(f \circ g)(2)$:
- First, find $g(2) = 2 + 3 = 5$
- Then, find $f(5) = 5^2 = 25$
- So $(f \circ g)(2) = 25$
To find the formula for $(f \circ g)(x)$:
- Start with $g(x) = x + 3$
- Plug that into $f$: $f(g(x)) = f(x + 3) = (x + 3)^2$
- So $(f \circ g)(x) = (x + 3)^2$
Important: Order matters! $(f \circ g)(x)$ is usually different from $(g \circ f)(x)$.
For the same $f$ and $g$ above:
- $(f \circ g)(x) = (x + 3)^2$
- $(g \circ f)(x) = g(f(x)) = g(x^2) = x^2 + 3$
These are completely different functions.
Inverse Functions: Undoing What a Function Does
An inverse function reverses what the original function does. If $f$ takes you from $a$ to $b$, then $f^{-1}$ (read “f inverse”) takes you from $b$ back to $a$.
More precisely: if $f(a) = b$, then $f^{-1}(b) = a$.
The notation $f^{-1}(x)$ is important: the $-1$ is NOT an exponent. This is not $\frac{1}{f(x)}$. The superscript $-1$ here specifically means “inverse function.”
The fundamental property of inverses:
- $f(f^{-1}(x)) = x$ for all $x$ in the domain of $f^{-1}$
- $f^{-1}(f(x)) = x$ for all $x$ in the domain of $f$
Composing a function with its inverse gets you back to where you started.
Example: If $f(x) = 2x$ (double the input), then $f^{-1}(x) = \frac{x}{2}$ (halve the input). Doubling then halving returns you to the original number.
Finding an inverse:
- Start with $y = f(x)$
- Swap $x$ and $y$ (interchange the roles of input and output)
- Solve for $y$
- The result is $f^{-1}(x)$
Not every function has an inverse. For a function to have an inverse, it must be one-to-one: each output corresponds to exactly one input. If two different inputs produce the same output, you cannot reverse the process unambiguously.
The Horizontal Line Test
The horizontal line test determines whether a function has an inverse.
Draw horizontal lines across the graph. If every horizontal line crosses the graph at most once, the function is one-to-one and has an inverse. If any horizontal line crosses more than once, the function is not one-to-one and does not have an inverse (unless you restrict the domain).
Example: $f(x) = x^2$ fails the horizontal line test. The line $y = 4$ crosses the parabola at both $x = 2$ and $x = -2$. Since two inputs give the same output, you cannot uniquely reverse the process.
Example: $f(x) = x^3$ passes the horizontal line test. Every horizontal line crosses the curve exactly once. This function has an inverse: $f^{-1}(x) = \sqrt[3]{x}$.
Notation and Terminology
| Term | Meaning | Example |
|---|---|---|
| $f(x)$ | Function notation: output of $f$ when input is $x$ | $f(x) = x^2 + 1$ |
| Domain | All valid inputs | ${x \mid x \neq 0}$ |
| Range | All possible outputs | ${y \mid y \geq 0}$ |
| $(f \circ g)(x)$ | Composition: f of g of x | If $f(x)=x^2$, $g(x)=x+1$, then $(f \circ g)(x)=(x+1)^2$ |
| $f^{-1}(x)$ | Inverse function (not $\frac{1}{f(x)}$!) | If $f(x)=2x$, then $f^{-1}(x)=\frac{x}{2}$ |
| Even function | Symmetric about y-axis: $f(-x)=f(x)$ | $f(x) = x^2$ |
| Odd function | Symmetric about origin: $f(-x)=-f(x)$ | $f(x) = x^3$ |
| Piecewise function | Different formulas for different intervals | See examples below |
| One-to-one | Each output comes from exactly one input | $f(x) = x^3$ |
Examples
Find the domain and range of $f(x) = \sqrt{x - 2}$.
Solution:
Finding the domain:
The square root function only works for non-negative inputs (in the real numbers). So we need whatever is under the radical to be greater than or equal to zero:
$$x - 2 \geq 0$$ $$x \geq 2$$
Domain: ${x \mid x \geq 2}$ or in interval notation: $[2, \infty)$
Finding the range:
The square root of any non-negative number is also non-negative. The smallest value under the radical is 0 (when $x = 2$), giving $\sqrt{0} = 0$. As $x$ increases, $\sqrt{x - 2}$ increases without bound.
Range: ${y \mid y \geq 0}$ or in interval notation: $[0, \infty)$
Verification: When $x = 2$, $f(2) = \sqrt{0} = 0$. When $x = 6$, $f(6) = \sqrt{4} = 2$. When $x = 11$, $f(11) = \sqrt{9} = 3$. The outputs are indeed non-negative and growing.
Determine whether $f(x) = x^4 - 3x^2$ is even, odd, or neither.
Solution:
To test, we compute $f(-x)$ and compare it to $f(x)$ and $-f(x)$.
Step 1: Find $f(-x)$ by replacing every $x$ with $-x$:
$$f(-x) = (-x)^4 - 3(-x)^2$$ $$f(-x) = x^4 - 3x^2$$
(Remember: $(-x)^4 = x^4$ because raising a negative to an even power gives a positive result. Similarly, $(-x)^2 = x^2$.)
Step 2: Compare to $f(x)$:
We have $f(-x) = x^4 - 3x^2$, which is exactly $f(x)$.
Since $f(-x) = f(x)$, the function is even.
Geometric interpretation: The graph is symmetric about the $y$-axis. If you fold the graph along the $y$-axis, the two halves match perfectly.
Note: We did not need to check if it is odd, because a function cannot be both even and odd (except for $f(x) = 0$).
Let $f(x) = 2x + 3$ and $g(x) = x^2$. Find $(f \circ g)(x)$ and $(g \circ f)(x)$.
Solution:
Finding $(f \circ g)(x)$:
This means $f(g(x))$. We take the output of $g$ and use it as the input to $f$.
Start with $g(x) = x^2$.
Now plug $x^2$ into $f$ wherever $f$ has an $x$:
$$f(g(x)) = f(x^2) = 2(x^2) + 3 = 2x^2 + 3$$
Answer: $(f \circ g)(x) = 2x^2 + 3$
Finding $(g \circ f)(x)$:
This means $g(f(x))$. We take the output of $f$ and use it as the input to $g$.
Start with $f(x) = 2x + 3$.
Now plug $2x + 3$ into $g$ wherever $g$ has an $x$:
$$g(f(x)) = g(2x + 3) = (2x + 3)^2$$
Expanding: $(2x + 3)^2 = 4x^2 + 12x + 9$
Answer: $(g \circ f)(x) = (2x + 3)^2 = 4x^2 + 12x + 9$
Key observation: $(f \circ g)(x) \neq (g \circ f)(x)$. The order of composition matters!
Verification with a specific value: Let $x = 1$.
- $(f \circ g)(1) = 2(1)^2 + 3 = 5$
- $(g \circ f)(1) = (2(1) + 3)^2 = 25$
These are indeed different.
Consider the piecewise function:
$$f(x) = \begin{cases} -x - 1 & \text{if } x < -1 \ x^2 & \text{if } -1 \leq x \leq 2 \ 4 & \text{if } x > 2 \end{cases}$$
Identify the domain, range, and where the function is increasing, decreasing, or constant.
Solution:
Domain:
The function has rules covering all real numbers:
- $x < -1$: covered by the first piece
- $-1 \leq x \leq 2$: covered by the second piece
- $x > 2$: covered by the third piece
Domain: All real numbers, $(-\infty, \infty)$
Evaluating at key points:
- At $x = -1$: We use the second rule (since $-1 \leq -1 \leq 2$): $f(-1) = (-1)^2 = 1$
- At $x = 0$: $f(0) = 0^2 = 0$
- At $x = 2$: $f(2) = 2^2 = 4$
- Just after $x = 2$: $f(2.1) = 4$ (from the third rule)
The function is continuous at $x = 2$ since both pieces give the value 4 there.
At $x = -1$: The first piece approaches $-(-1) - 1 = 0$ as $x$ approaches $-1$ from the left, but $f(-1) = 1$. There is a jump discontinuity at $x = -1$.
Range:
- For $x < -1$: The line $-x - 1$ gives outputs greater than 0 (approaching but never reaching 0)
- For $-1 \leq x \leq 2$: The parabola $x^2$ gives outputs from 0 (at $x = 0$) to 4 (at $x = 2$), including 1 (at $x = -1$)
- For $x > 2$: The constant 4
The smallest output is 0 (at $x = 0$). Outputs include everything from 0 to 4 on the parabola, plus the value 4 extending to the right, plus all values greater than 0 from the line on the left.
Range: $[0, \infty)$
Behavior:
- Decreasing on $(-\infty, -1)$: The line $-x - 1$ goes downhill to the right
- Decreasing on $(-1, 0)$: The parabola $x^2$ goes downhill toward its vertex
- Increasing on $(0, 2)$: The parabola $x^2$ goes uphill from its vertex
- Constant on $(2, \infty)$: The function stays at 4
Minimum: Absolute minimum of 0 at $x = 0$ (the vertex of the parabola)
Maximum: No absolute maximum (but the function never exceeds 4 for $x \geq -1$; however, for $x < -1$, the linear piece extends upward without bound)
Wait, let me reconsider. For $x < -1$, the piece is $-x - 1$. When $x = -2$, $f(-2) = -(-2) - 1 = 1$. When $x = -10$, $f(-10) = -(-10) - 1 = 9$. So this piece increases as $x$ decreases (equivalently, decreases as $x$ increases from left to right).
Actually, let me recalculate the behavior of the first piece. As $x$ increases toward $-1$ from the left, $-x$ decreases, so $-x - 1$ decreases. At $x = -3$, $f(-3) = 3 - 1 = 2$. At $x = -2$, $f(-2) = 2 - 1 = 1$. At $x \to -1^-$, $f(x) \to 1 - 1 = 0$. So the first piece is indeed decreasing as $x$ increases.
Corrected behavior:
- Decreasing on $(-\infty, -1)$
- Decreasing on $[-1, 0]$
- Increasing on $[0, 2]$
- Constant on $[2, \infty)$
The function has no absolute maximum since the line $-x - 1$ grows without bound as $x \to -\infty$.
Find the inverse of $f(x) = \frac{2x + 1}{x - 3}$.
Solution:
Step 1: Write $y = f(x)$: $$y = \frac{2x + 1}{x - 3}$$
Step 2: Swap $x$ and $y$: $$x = \frac{2y + 1}{y - 3}$$
Step 3: Solve for $y$:
Multiply both sides by $(y - 3)$: $$x(y - 3) = 2y + 1$$
Distribute: $$xy - 3x = 2y + 1$$
Get all $y$ terms on one side: $$xy - 2y = 3x + 1$$
Factor out $y$: $$y(x - 2) = 3x + 1$$
Divide by $(x - 2)$: $$y = \frac{3x + 1}{x - 2}$$
Step 4: Write the inverse: $$f^{-1}(x) = \frac{3x + 1}{x - 2}$$
Verification:
Let us check that $f(f^{-1}(x)) = x$:
$$f(f^{-1}(x)) = f\left(\frac{3x + 1}{x - 2}\right) = \frac{2 \cdot \frac{3x + 1}{x - 2} + 1}{\frac{3x + 1}{x - 2} - 3}$$
Simplify the numerator: $$2 \cdot \frac{3x + 1}{x - 2} + 1 = \frac{2(3x + 1)}{x - 2} + \frac{x - 2}{x - 2} = \frac{6x + 2 + x - 2}{x - 2} = \frac{7x}{x - 2}$$
Simplify the denominator: $$\frac{3x + 1}{x - 2} - 3 = \frac{3x + 1}{x - 2} - \frac{3(x - 2)}{x - 2} = \frac{3x + 1 - 3x + 6}{x - 2} = \frac{7}{x - 2}$$
So: $$f(f^{-1}(x)) = \frac{\frac{7x}{x - 2}}{\frac{7}{x - 2}} = \frac{7x}{x - 2} \cdot \frac{x - 2}{7} = x$$
The inverse is verified.
Domain considerations: The original function $f(x)$ has domain $x \neq 3$. The inverse $f^{-1}(x)$ has domain $x \neq 2$.
Verify that $f(x) = 3x - 7$ and $g(x) = \frac{x + 7}{3}$ are inverse functions.
Solution:
Two functions are inverses if and only if both of these are true:
- $f(g(x)) = x$ for all $x$ in the domain of $g$
- $g(f(x)) = x$ for all $x$ in the domain of $f$
Check 1: $f(g(x)) = x$
$$f(g(x)) = f\left(\frac{x + 7}{3}\right)$$
Substitute $\frac{x + 7}{3}$ into $f$: $$= 3 \cdot \frac{x + 7}{3} - 7$$ $$= (x + 7) - 7$$ $$= x$$
Check 2: $g(f(x)) = x$
$$g(f(x)) = g(3x - 7)$$
Substitute $3x - 7$ into $g$: $$= \frac{(3x - 7) + 7}{3}$$ $$= \frac{3x}{3}$$ $$= x$$
Both conditions are satisfied, so $f$ and $g$ are indeed inverse functions.
Interpretation: The function $f(x) = 3x - 7$ says “multiply by 3, then subtract 7.” The inverse $g(x) = \frac{x + 7}{3}$ says “add 7, then divide by 3.” Notice that the inverse undoes the operations in reverse order.
This is like putting on socks and shoes. You put on socks first, then shoes. To undo it, you take off shoes first, then socks. Functions work the same way.
Key Properties and Rules
Domain Restrictions to Remember
| Expression Type | Restriction | Example |
|---|---|---|
| Division | Denominator $\neq 0$ | $\frac{1}{x-3}$: domain excludes $x = 3$ |
| Square root (even roots) | Radicand $\geq 0$ | $\sqrt{x+5}$: domain is $x \geq -5$ |
| Logarithm | Argument $> 0$ | $\ln(x-2)$: domain is $x > 2$ |
Tests and What They Tell You
| Test | What It Checks | Pass Means | Fail Means |
|---|---|---|---|
| Vertical Line Test | Is it a function? | Each $x$ gives one $y$ | Not a function |
| Horizontal Line Test | Is it one-to-one? | Has an inverse | No inverse (without restricting domain) |
Even and Odd Functions: Quick Reference
| Type | Algebraic Test | Graphical Symmetry | Examples |
|---|---|---|---|
| Even | $f(-x) = f(x)$ | Symmetric about $y$-axis | $x^2$, $x^4$, $ |
| Odd | $f(-x) = -f(x)$ | Symmetric about origin | $x$, $x^3$, $x^5$, $\sin(x)$ |
| Neither | Neither equation holds | No special symmetry | $x^2 + x$, $e^x$, most functions |
Composition: Order Matters
- $(f \circ g)(x) = f(g(x))$: Do $g$ first, then $f$
- $(g \circ f)(x) = g(f(x))$: Do $f$ first, then $g$
- These are usually different!
- Read inside-out, evaluate inside-out
Inverse Functions: Key Facts
- Notation: $f^{-1}(x)$ means inverse, NOT $\frac{1}{f(x)}$
- $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$
- The domain of $f$ becomes the range of $f^{-1}$
- The range of $f$ becomes the domain of $f^{-1}$
- Graphically: $f$ and $f^{-1}$ are reflections over the line $y = x$
Real-World Applications
Temperature Conversion and Its Inverse
The formula to convert Celsius to Fahrenheit is: $$F(C) = \frac{9}{5}C + 32$$
To convert Fahrenheit back to Celsius, we need the inverse: $$C(F) = \frac{5}{9}(F - 32)$$
These are inverse functions. If you convert 20 degrees Celsius to Fahrenheit (getting 68), then convert 68 Fahrenheit back to Celsius, you get 20 again. The functions undo each other.
Encoding and Decoding Messages
Cryptography uses inverse functions. A simple encoding function might shift each letter by 3 positions: A becomes D, B becomes E, and so on. The decoding function shifts back by 3 positions: D becomes A, E becomes B. Encoding and decoding are inverse operations.
Modern encryption uses much more sophisticated functions, but the principle remains: there is an encoding function and a decoding function (the inverse), and only someone with the inverse can read the message.
Compound Processes (Composition)
Many real processes involve applying one function, then another.
Example: A store offers a 20% discount, and you have a coupon for $10 off. Let $d(x) = 0.8x$ be the discount function and $c(x) = x - 10$ be the coupon function.
- If the discount is applied first, then the coupon: $(c \circ d)(x) = c(0.8x) = 0.8x - 10$
- If the coupon is applied first, then the discount: $(d \circ c)(x) = d(x - 10) = 0.8(x - 10) = 0.8x - 8$
On a $100 item:
- Discount first, then coupon: $0.8(100) - 10 = 80 - 10 = $70
- Coupon first, then discount: $0.8(100 - 10) = 0.8(90) = $72
The order matters! This is composition in action.
Piecewise Functions in Daily Life
Tax brackets are the classic example. Your income is not taxed at a single rate. Instead, different portions are taxed at different rates.
For a simplified example, suppose:
- Income up to $10,000 is taxed at 10%
- Income from $10,001 to $40,000 is taxed at 20%
- Income above $40,000 is taxed at 30%
The tax function $T(x)$ on income $x$ would be piecewise, with different formulas for each bracket.
Shipping costs often work similarly. A shipping company might charge:
- $5 for packages up to 1 lb
- $8 for packages from 1 to 5 lbs
- $8 plus $2 per pound over 5 lbs for heavier packages
This is a piecewise function of weight.
Self-Test Problems
Problem 1: Find the domain and range of $f(x) = \frac{1}{x + 4}$.
Show Answer
Domain: The denominator cannot be zero, so $x + 4 \neq 0$, meaning $x \neq -4$.
Domain: All real numbers except $x = -4$, written as ${x \mid x \neq -4}$ or $(-\infty, -4) \cup (-4, \infty)$.
Range: As $x$ approaches $-4$ from either side, $\frac{1}{x+4}$ approaches $\pm\infty$. As $x$ approaches $\pm\infty$, $\frac{1}{x+4}$ approaches 0 but never equals 0.
Range: All real numbers except $y = 0$, written as ${y \mid y \neq 0}$ or $(-\infty, 0) \cup (0, \infty)$.
Problem 2: Determine whether $f(x) = x^3 + x$ is even, odd, or neither.
Show Answer
Calculate $f(-x)$: $$f(-x) = (-x)^3 + (-x) = -x^3 - x = -(x^3 + x) = -f(x)$$
Since $f(-x) = -f(x)$, the function is odd.
The graph has rotational symmetry about the origin.
Problem 3: Given $f(x) = x - 5$ and $g(x) = x^2 + 1$, find $(f \circ g)(3)$ and $(g \circ f)(3)$.
Show Answer
$(f \circ g)(3)$: First, $g(3) = 3^2 + 1 = 10$. Then, $f(10) = 10 - 5 = 5$. Answer: $(f \circ g)(3) = 5$
$(g \circ f)(3)$: First, $f(3) = 3 - 5 = -2$. Then, $g(-2) = (-2)^2 + 1 = 5$. Answer: $(g \circ f)(3) = 5$
(In this case, they happen to give the same value for $x = 3$, but the functions $(f \circ g)(x) = x^2 - 4$ and $(g \circ f)(x) = (x-5)^2 + 1 = x^2 - 10x + 26$ are different.)
Problem 4: Find the inverse of $f(x) = 5x + 2$.
Show Answer
Step 1: Write $y = 5x + 2$
Step 2: Swap $x$ and $y$: $x = 5y + 2$
Step 3: Solve for $y$: $$x - 2 = 5y$$ $$y = \frac{x - 2}{5}$$
Answer: $f^{-1}(x) = \frac{x - 2}{5}$
Verification: $f(f^{-1}(x)) = 5 \cdot \frac{x-2}{5} + 2 = (x - 2) + 2 = x$
Problem 5: Evaluate the piecewise function at $x = -2$, $x = 0$, and $x = 3$:
$$f(x) = \begin{cases} x + 4 & \text{if } x < 0 \ x^2 - 1 & \text{if } x \geq 0 \end{cases}$$
Show Answer
At $x = -2$: Since $-2 < 0$, use the first rule. $f(-2) = (-2) + 4 = 2$
At $x = 0$: Since $0 \geq 0$, use the second rule. $f(0) = (0)^2 - 1 = -1$
At $x = 3$: Since $3 \geq 0$, use the second rule. $f(3) = (3)^2 - 1 = 8$
Problem 6: Does $f(x) = x^2 + 1$ have an inverse? Explain using the horizontal line test.
Show Answer
No, $f(x) = x^2 + 1$ does not have an inverse (without restricting the domain).
Horizontal line test: The graph is a parabola opening upward with vertex at $(0, 1)$. Any horizontal line above $y = 1$ (such as $y = 5$) crosses the parabola at two points.
For example, if $y = 5$, then $x^2 + 1 = 5$, so $x^2 = 4$, giving $x = 2$ or $x = -2$. Two different inputs produce the same output.
Since the function is not one-to-one, it has no inverse.
(However, if we restrict the domain to $x \geq 0$, the function becomes one-to-one and the inverse would be $f^{-1}(x) = \sqrt{x - 1}$.)
Problem 7: If $f(x) = \frac{x}{x+1}$, find $f(f(x))$.
Show Answer
We need $f(f(x))$, which means we substitute $f(x)$ into $f$.
First, $f(x) = \frac{x}{x+1}$.
Now substitute this into $f$:
$$f(f(x)) = f\left(\frac{x}{x+1}\right) = \frac{\frac{x}{x+1}}{\frac{x}{x+1} + 1}$$
Simplify the denominator: $$\frac{x}{x+1} + 1 = \frac{x}{x+1} + \frac{x+1}{x+1} = \frac{x + x + 1}{x+1} = \frac{2x+1}{x+1}$$
So: $$f(f(x)) = \frac{\frac{x}{x+1}}{\frac{2x+1}{x+1}} = \frac{x}{x+1} \cdot \frac{x+1}{2x+1} = \frac{x}{2x+1}$$
Answer: $f(f(x)) = \frac{x}{2x+1}$
Summary
- A function assigns exactly one output to each input. This predictability is what makes functions so useful.
- Domain is all valid inputs; range is all possible outputs. Know the common restrictions: no division by zero, no square roots of negatives (in reals), no logarithms of non-positives.
- Functions can be increasing (going up), decreasing (going down), or constant (flat) on different intervals. We describe these using $x$-intervals.
- Local maxima and minima are high and low points relative to nearby values. Absolute maxima and minima are the highest and lowest points overall.
- Even functions satisfy $f(-x) = f(x)$ and have $y$-axis symmetry. Odd functions satisfy $f(-x) = -f(x)$ and have origin symmetry. Most functions are neither.
- Piecewise functions use different rules for different parts of the domain. They model real situations like tax brackets and shipping costs.
- Composition $(f \circ g)(x) = f(g(x))$ applies one function to the output of another. Order matters: $(f \circ g)(x)$ is usually different from $(g \circ f)(x)$.
- An inverse function $f^{-1}$ undoes what $f$ does: $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$.
- The horizontal line test determines if a function has an inverse. If any horizontal line crosses the graph more than once, no inverse exists (without restricting the domain).
- The notation $f^{-1}(x)$ means inverse function, not $\frac{1}{f(x)}$.
- Inverse functions swap domain and range. Their graphs are reflections over the line $y = x$.
- Functions and their properties are foundational to everything in Algebra 2 and beyond. Understanding them deeply now will make every subsequent topic easier.