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Special and piecewise functions

Piecewise Functions: Graphs, Limits, Continuity and Derivatives

Piecewise functions explained: how to read and graph them, find one-sided limits, check continuity and differentiability, and integrate across pieces.

Special and piecewise functions Updated 3 min read

At a glance

f(x) = \begin{cases} f_1(x) & x < a \\ f_2(x) & x \ge a \end{cases}
Domain
The union of the intervals the pieces cover
Continuous at x = a
When \lim_{x\to a^-}f_1(x) = \lim_{x\to a^+}f_2(x) = f(a)
Derivative
f'(x) = \begin{cases} f_1'(x) & x < a \\ f_2'(x) & x > a \end{cases}
Integral
\int_b^c f(x)\,dx = \int_b^a f_1(x)\,dx + \int_a^c f_2(x)\,dx
Examples
Absolute value, floor function, tax brackets, shipping rates
On this page
  1. How to read a piecewise function
  2. Graph of a piecewise function
  3. Limits of piecewise functions
  4. Continuity at a break point
  5. Derivatives of piecewise functions
  6. Integrating a piecewise function
  7. Common mistakes
  8. Related functions

A piecewise function is a function defined by different formulas on different parts of its domain. The rule tells you which formula to use based on where x is. Piecewise functions are the standard test cases for limits, continuity and differentiability, because the interesting behavior happens where the pieces meet.

How to read a piecewise function

f(x) = \begin{cases} x^2 & x < 1 \\ 3 - x & x \ge 1 \end{cases}

To evaluate, first find which condition x satisfies, then use that piece only. Here f(0) = 0^2 = 0, f(1) = 3 - 1 = 2 (because x \ge 1 includes 1), and f(2) = 3 - 2 = 1.

Graph of a piecewise function

Graph of a piecewise function with a jump at x = 1
The left piece x^2 approaches the open dot at (1, 1). The right piece 3 - x starts at the closed dot (1, 2), which is the actual value f(1).

Each piece is drawn only on its own interval. An open dot marks an endpoint that is not included; a closed dot marks one that is.

Limits of piecewise functions

At a break point a, compute the two one-sided limits separately, each with its own piece. For the function above at a = 1:

\lim_{x\to 1^-}f(x) = \lim_{x\to 1^-}x^2 = 1, \qquad \lim_{x\to 1^+}f(x) = \lim_{x\to 1^+}(3 - x) = 2

The one-sided limits differ, so \lim_{x\to 1}f(x) does not exist, and the graph has a jump. Away from the break point, a limit is just direct substitution into the active piece.

Continuity at a break point

f is continuous at a when all three of these agree:

  1. the left-hand limit \lim_{x\to a^-}f(x),
  2. the right-hand limit \lim_{x\to a^+}f(x),
  3. the value f(a).

Example: choosing a constant. For which k is

g(x) = \begin{cases} kx + 1 & x < 2 \\ x^2 & x \ge 2 \end{cases}

continuous? The left limit at 2 is 2k + 1 and the right limit and value are 2^2 = 4. Setting 2k + 1 = 4 gives k = \frac{3}{2}.

Derivatives of piecewise functions

Inside each open interval, differentiate the active piece as usual. At a break point you need two checks:

  • f must be continuous there (a jump means no derivative).
  • The left and right derivatives must match: \lim_{h\to 0^-}\frac{f(a+h) - f(a)}{h} = \lim_{h\to 0^+}\frac{f(a+h) - f(a)}{h}.

When both pieces are differentiable polynomials and f is continuous at a, the second check reduces to comparing f_1'(a) with f_2'(a).

Example: a corner. The absolute value function |x| is -x for x < 0 and x for x \ge 0. It is continuous at 0, but the left slope is -1 and the right slope is 1, so |x| is not differentiable at 0.

Example: continuous but not smooth. For g above with k = \frac{3}{2}, the left slope at 2 is \frac{3}{2} and the right slope is 2x = 4. They differ, so g is continuous but not differentiable at x = 2.

Integrating a piecewise function

Split the integral at every break point inside the interval. For the first function on [0, 3]:

\int_0^3 f(x)\,dx = \int_0^1 x^2\,dx + \int_1^3 (3 - x)\,dx = \frac{1}{3} + 2 = \frac{7}{3}

The second integral is \left[3x - \frac{x^2}{2}\right]_1^3 = \frac{9}{2} - \frac{5}{2} = 2. A single jump does not change the value of the integral, so it does not matter which piece owns the point x = 1.

Common mistakes

  1. Using the wrong piece at the boundary. Check whether the condition says < or \le before evaluating f(a).
  2. Concluding continuity from one side. Both one-sided limits and the value must agree.
  3. Assuming continuous means differentiable. Corners like |x| at 0 are continuous but have no derivative.
  4. Differentiating the pieces and comparing them at a jump. If f is not continuous at a, it is not differentiable there, even if the piece slopes happen to match.
  5. Integrating one formula across the whole interval. Always split at the break points.

The floor and ceiling functions are piecewise constant step functions with a jump at every integer, and absolute value is the simplest piecewise function with a corner. For the full treatment of limits and continuity, see the guide to continuity.

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Frequently asked questions

How do you find the limit of a piecewise function?

Away from a break point, use the formula for that piece. At a break point, compute the left-hand limit with the left piece and the right-hand limit with the right piece. The limit exists only if they agree.

How do you make a piecewise function continuous?

At each break point a, set the two pieces equal at x = a and solve for the unknown constant. This forces the left and right limits to match f(a).

Is a continuous piecewise function always differentiable?

No. The pieces can meet with different slopes, creating a corner. The absolute value function |x| is continuous at 0 but not differentiable there.

How do you graph a piecewise function?

Graph each formula only on its own interval. At each endpoint draw a closed dot if the endpoint is included and an open dot if it is not.

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