The Twelve Basic Functions

Ever wondered why some functions are considered "basic" in math? These twelve functions form the foundation of more complex mathematics. Let's explore their essential characteristics!

Each function has specific properties that make it unique. For the identity function $f(x) = x$, the input equals the output, creating a straight line through the origin with domain and range of all real numbers. The squaring function $f(x) = x²$ creates that familiar U-shaped parabola, while the exponential function $f(x) = eˣ$ grows increasingly rapidly.

Some functions have interesting behaviors at their edges. The natural logarithm function (ln(x)) has a vertical asymptote at x = 0, meaning it approaches negative infinity as x approaches zero. The reciprocal function $1/x$ has both horizontal and vertical asymptotes, creating a hyperbola.

Functions like sine and cosine oscillate between -1 and 1, creating those wave patterns you see everywhere from sound waves to electricity. The greatest integer function creates a staircase pattern because it rounds down to the nearest integer.

💡 Quick Tip: Pay attention to each function's domain, range, and symmetry. These properties will help you recognize the functions and sketch them quickly on tests!

Greatest Integer Function & Building Functions

The Greatest Integer Function is a bit tricky but super useful! It rounds numbers down to the nearest integer. Your textbook notation is int(x), but you might also see it as [x] in other materials.

Think of the Greatest Integer Function as always rounding down, no matter how close a number is to the next integer. For example, int(3.999) = 3, and int(4.001) = 4. With negative numbers, you still round down, so int(-2.1) = -3 $not -2$. When graphed, this function creates a staircase pattern with open circles at one end of each "step."

Building composite functions means plugging one function into another. If f(x) = 2x-3 and you're asked to find f(2), simply replace x with 2: f(2) = 2(2)-3 = 4-3 = 1. For something like f$3x-1$, substitute 3x-1 wherever you see x: f$3x-1$ = 2$3x-1$-3 = 6x-2-3 = 6x-5.

When working with two different functions like f(z) = -z + 4 and h(z) = 2z, finding f(h(3)) means you first calculate h(3) = 2(3) = 6, then find f(6) = -6 + 4 = -2. We write this as (f∘h)(3) = -2, which is read as "f composed with h of 3."

🔑 Remember: Always work from inside the parentheses outward! The innermost function gets evaluated first, then its result becomes the input for the outer function.

Composition of Functions

Composition functions might look intimidating, but they're actually just functions working together! When we write (f∘g)(x), we're simply saying "apply function g first, then apply function f to that result."

Let's see how this works. If f(x) = x + 1 and g(x) = x², then f(g(x)) means f(x²) = x² + 1. We've just created a new function by composition! Similarly, if f(x) = 3x + 5 and g(x) = x - 2, then (f∘g)(x) = f(g(x)) = f$x-2$ = 3$x-2$ + 5 = 3x - 6 + 5 = 3x - 1.

Notice that (f∘g)(x) and (g∘f)(x) usually give different results. For example, with h(x) = x² - 1 and p(x) = 2x, we get (h∘p)(x) = h(p(x)) = h(2x) = (2x)² - 1 = 4x² - 1, while (p∘h)(x) = p(h(x)) = p$x² - 1$ = 2$x² - 1$ = 2x² - 2. The order matters!

When finding domains of composite functions, you need to consider two things: the domain of the inner function AND what values from that domain will work in the outer function. This gets especially important with fractions (where denominators can't be zero) and square roots (where expressions under the radical can't be negative).

🧠 Pro Tip: To find the domain of f(g(x)), first identify what values of x work in g(x), then check which of those outputs from g will work as inputs for f(x). The final domain is the intersection of these constraints.

Domains and Decomposing Functions

Finding domains of composite functions requires careful attention to restrictions. Different types of functions have different domain limitations that you need to watch for.

Polynomial functions $like x² + 3x + 1$ have domains of all real numbers, but fractions require that denominators never equal zero. Square roots need whatever is inside the radical to be non-negative. When functions are combined, you need to check all restrictions from both functions.

For example, if f(x) = x² + 1 and g(x) = √x, then the domain of g(x) is [0,∞) because you can't take the square root of a negative number. When finding f(g(x)) = (√x)² + 1 = x + 1, the domain is still [0,∞) because that's what g(x) requires as input.

Decomposing functions means working backward from a composite function to find its components. When given f(g(x)), you need to identify what the "mother function" f(x) is, and then determine what was plugged into it (g(x)). For example, if f(g(x)) = √x - 1, the mother function might be f(x) = √x, and g(x) = x - 1.

There's often more than one correct way to decompose a function. For f(g(x)) = 2/$3x+1$, you might identify f(x) = 2/x as the mother function, making g(x) = 3x+1. This is like detective work - you're looking for clues about what operations were performed in what order.

💡 Quick Check: To verify your decomposition, plug g(x) into f(x) and make sure you get the original composite function. If f(x) = 2/x and g(x) = 3x+1, then f(g(x)) = 2/$3x+1$, which matches!

Inverse Functions

Inverse functions essentially "undo" what the original function does. If a function takes x to y, its inverse takes y back to x. But not all functions have inverses!

How can you tell if a function has an inverse? Use the horizontal line test. If any horizontal line intersects your function's graph more than once, the function doesn't have an inverse. This makes sense because an inverse function needs to "know" exactly which x-value to return to, with no ambiguity.

A one-to-one function passes both the vertical line test (making it a function) and the horizontal line test (meaning it has an inverse). Functions that fail the horizontal line test, like y = x², don't have inverses because multiple inputs give the same output $like both 2 and -2 give 4$.

To find an inverse algebraically, follow these steps: Replace f(x) with y, swap x and y, solve for y, and then replace y with f⁻¹(x). For example, to find the inverse of f(x) = $x+2$/x:

1. Write y = $x+2$/x
2. Multiply both sides by x: xy = x+2
3. Rearrange: xy - x = 2
4. Factor out x: x$y-1$ = 2
5. Solve for x: x = 2/$y-1$
6. Replace y with x: f⁻¹(x) = 2/$x-1$

⚡ Power Move: When graphing a function and its inverse, they're always reflections of each other across the line y = x. This is a great way to check if you've found the correct inverse!

Practice with Composite Functions

Now it's time to put your skills to work with some practice problems! Working through these examples will help cement your understanding of composite functions.

When computing composite functions like f$g(-3)$, always work from the inside out. First calculate g(-3) = 3(-3) = -9, then find f(-9) = 2(-9) - 1 = -18 - 1 = -19. For more complex compositions like f(g(h(2))), work step by step: h(2) = 2² + 1 = 5, then g(5) = 3(5) = 15, and finally f(15) = 2(15) - 1 = 30 - 1 = 29.

Composite functions can also involve expressions rather than specific values. For f$x + 1$, substitute x + 1 wherever you see x in the original function: f$x + 1$ = 2$x + 1$ - 1 = 2x + 2 - 1 = 2x + 1. Similarly, h$x - 2$ = $x - 2$² + 1 = x² - 4x + 4 + 1 = x² - 4x + 5.

When finding composite functions with algebraic expressions like f(g(x)), substitute the entire g(x) expression wherever x appears in f(x). If f(x) = -3x + 7 and g(x) = 2x² - 8, then f(g(x)) = -3$2x² - 8$ + 7 = -6x² + 24 + 7 = -6x² + 31.

🏆 Challenge Yourself: Try making up your own composite function problems and solving them. The more practice you get, the more confident you'll feel on tests!