Fundamentals of Piecewise-Defined Functions

TL;DR

A piecewise function uses different formulas on different parts of the domain, like switching rules depending on where x lands. You'll learn how to evaluate them, graph them piece by piece, and check whether they're continuous at the breakpoints. By the end, you'll read piecewise notation fluently and spot discontinuities without plotting every point.

1. The Mental Model

Think of a piecewise function as a set of rules with conditions attached, like a tax bracket system: which formula applies depends entirely on which bracket (interval) your input falls into. You never use two formulas at once — you pick exactly one, based on the x-value. A piecewise function is just several ordinary functions stitched together, each one only "awake" on its own slice of the domain.

2. The Core Material

2.1 Reading and Evaluating Piecewise Notation

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A piecewise function is written as a list of formulas, each paired with a condition on x. Here's a typical example:

$$ f(x) = \begin{cases} x^2 & \text{if } x < 0 \\ 2x + 1 & \text{if } 0 \le x < 3 \\ 7 & \text{if } x \ge 3 \end{cases} $$

To evaluate $f(x)$ at a specific input, you do exactly one thing: find which condition the number satisfies, then plug it into that formula only. Ignore the others completely.

• $f(-2)$: is $-2 < 0$? Yes. Use $x^2$: $f(-2) = (-2)^2 = 4$.
• $f(0)$: is $0 < 0$? No. Is $0 \le 0 < 3$? Yes. Use $2x+1$: $f(0) = 2(0)+1 = 1$.
• $f(5)$: is $5 \ge 3$? Yes. Use the constant: $f(5) = 7$.

The single biggest source of errors here is sloppiness about strict vs. non-strict inequalities ($<$ vs. $\le$). Every x-value in the domain must belong to exactly one interval — never zero, never two. If you ever find a number satisfying two conditions, the function is badly defined (or you misread it).

2.2 Graphing Piece by Piece

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Graphing a piecewise function means graphing each formula, but only drawing the portion that lies inside its own interval — then erasing (mentally) everything outside that interval.

Practical method:
1. Graph each piece as if it were the whole function, lightly.
2. Trim each graph to only the x-values allowed by its condition.
3. Mark the boundary points carefully: a closed dot (●) means that exact point is included (the inequality has $\le$ or $\ge$); an open dot (○) means it's excluded (strict $<$ or $>$).
4. Check each breakpoint (x = 0 and x = 3 in our example) — there should be exactly one dot there, not two closed or two open dots.

For our function above:
- On $x<0$: part of the parabola $y=x^2$, approaching $(0,0)$ with an open circle (since $0$ is not included here).
- On $0 \le x < 3$: part of the line $y=2x+1$, starting with a closed dot at $(0,1)$ and an open dot at $(3,7)$.
- On $x \ge 3$: a horizontal ray $y=7$ starting with a closed dot at $(3,7)$.

Notice $(0,0)$ from the parabola piece is open, and $(0,1)$ from the line piece is closed — these don't collide because they're different y-values at the same x. That's fine; it just means there's a jump at $x=0$.

flowchart TD
    A["Pick x-value"] --> B{"x < 0?"}
    B -- Yes --> C["Use f(x) = x^2"]
    B -- No --> D{"0 <= x < 3?"}
    D -- Yes --> E["Use f(x) = 2x + 1"]
    D -- No --> F{"x >= 3?"}
    F -- Yes --> G["Use f(x) = 7"]

2.3 Continuity at the Breakpoints

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A breakpoint is where the rule switches (here, $x=0$ and $x=3$). The function might connect smoothly there, or it might jump. To check, compare three things at each breakpoint $x=a$:

1. The left-hand limit: plug values slightly less than $a$ into the piece that applies there.
2. The right-hand limit: plug values slightly more than $a$ into the piece that applies there.
3. The actual function value $f(a)$, using whichever piece's condition includes $a$ itself.

The function is continuous at $x=a$ only if all three agree:
$$\lim_{x\to a^-} f(x) = \lim_{x\to a^+} f(x) = f(a)$$

Check $x=0$ for our function:
- Left piece $x^2$ as $x\to 0^-$: gives $0^2 = 0$.
- Right piece $2x+1$ as $x\to 0^+$: gives $2(0)+1=1$.
- $f(0)$ itself (using the second piece, since $0$ satisfies $0 \le x <3$): $f(0)=1$.

Left limit ($0$) ≠ right limit ($1$), so there's a jump discontinuity at $x=0$, even though $f(0)$ matches the right side.

Check $x=3$:
- Left piece $2x+1$ as $x\to 3^-$: $2(3)+1=7$.
- Right piece (constant $7$) as $x\to 3^+$: $7$.
- $f(3)$ itself (third piece includes $3$): $7$.

All three equal $7$ — continuous at $x=3$. The graph connects smoothly there with no visible break, even though the formula changes.

This three-way check is the whole game. Most mistakes come from using the wrong piece when computing $f(a)$ — always use the piece whose inequality includes the exact boundary point.

3. Worked Example

Let's fully analyze this function:

$$ g(x) = \begin{cases} -x + 2 & \text{if } x \le 1 \\ x^2 & \text{if } 1 < x < 4 \\ 3x - 8 & \text{if } x \ge 4 \end{cases} $$

Step 1 — Evaluate at sample points.
- $g(-3)$: satisfies $x \le 1$, so use $-x+2$: $g(-3) = -(-3)+2 = 5$.
- $g(2)$: satisfies $1 < x < 4$, so use $x^2$: $g(2) = 4$.
- $g(4)$: satisfies $x \ge 4$, so use $3x-8$: $g(4) = 3(4)-8 = 4$.

Step 2 — Graph each piece, respecting domain restrictions.
- Piece 1 ($-x+2$) is a line valid for all $x \le 1$. At $x=1$: $g(1) = -1+2 = 1$, and since $1\le1$ is true, this point gets a closed dot at $(1,1)$.
- Piece 2 ($x^2$) is valid for $1 < x < 4$ only. As $x\to 1^+$, $x^2 \to 1$, so there's an open dot at $(1,1)$ (same location as the closed dot from piece 1 — good sign). As $x \to 4^-$, $x^2 \to 16$, giving an open dot at $(4,16)$.
- Piece 3 ($3x-8$) is valid for $x \ge 4$. At $x=4$: $3(4)-8=4$, so there's a closed dot at $(4,4)$ — note this is not at $(4,16)$.

Step 3 — Check continuity at $x=1$.
- Left limit (piece 1, $x\to 1^-$): $-x+2 \to 1$.
- Right limit (piece 2, $x\to 1^+$): $x^2 \to 1$.
- $g(1)$ (piece 1 applies since $1 \le 1$): $g(1) = 1$.

All three equal $1$. Continuous at $x=1$. The open dot from piece 2 and the closed dot from piece 1 sit exactly on top of each other — the graph connects without a jump.

Step 4 — Check continuity at $x=4$.
- Left limit (piece 2, $x\to 4^-$): $x^2 \to 16$.
- Right limit (piece 3, $x\to 4^+$): $3x-8 \to 4$.
- $g(4)$ (piece 3 applies since $4 \ge 4$): $g(4) = 4$.

Left limit ($16$) ≠ right limit ($4$). Discontinuous (jump) at $x=4$. The graph drops suddenly from near $y=16$ down to $y=4$ as it crosses $x=4$. The open dot at $(4,16)$ from the parabola piece marks where the graph "would have gone" if it kept following piece 2, but the actual function jumps down to the closed dot at $(4,4)$ instead.

This single example shows both outcomes: one smooth join and one real jump, which is exactly the kind of mixed behavior piecewise functions are known for.

4. Key Takeaways