Making a Piecewise Function Continuous

A jump at the boundary

Take f(x) = 2x for x < 2, and f(x) = x + 4 for $x\ge 2$x ≥ 2. As x comes up to 2 from the left, 2x comes closer and closer to 2 × 2 = 4, so the left piece ends in a hollow dot at (2, 4). The right piece starts at x = 2 with 2 + 4 = 6, a filled dot at (2, 6). The graph jumps by 6 − 4 = 2 at the boundary.

A graph that can be drawn without lifting the pen is called continuous. This one is not continuous at x = 2: the pen has to jump from the end of one piece to the start of the other.

f(x) = 2x for x < 2 and x + 4 for $x\ge 2$x ≥ 2. The left piece ends in a hollow dot at (2, 4) and the right piece starts from a filled dot at (2, 6): a jump of 2.

Leave the constant unknown

Replace the 4 in the second rule by an unknown constant k: f(x) = 2x for x < 2, and f(x) = x + k for $x\ge 2$x ≥ 2. Changing k slides the whole right-hand piece up or down, without changing its gradient. The question is which value of k makes the right-hand piece start exactly where the left-hand piece ends.

Work out both rules at the boundary, x = 2. The left rule heads for 2 × 2 = 4. The right rule gives 2 + k.

Set the two values equal

The pieces meet when the two values are the same, so set them equal: 2 + k = 4. Subtract 2 from both sides: k = 2. The equation has one solution, so exactly one value of k closes the jump.

Check: with k = 2 the right rule is x + 2, and at x = 2 it gives 2 + 2 = 4, the same as the left rule. The filled dot of the right-hand piece now sits on the hollow end of the left-hand piece, at (2, 4), and fills it.

The gold graph is f with k = 2: both pieces pass through (2, 4). The white lines are the right-hand piece with k = 4, which starts at (2, 6), too high, and with k = 0, which starts at (2, 2), too low.

Continuous at the boundary

With k = 2 the graph runs through x = 2 unbroken, so f is continuous there. Within each piece nothing breaks, because each rule is a straight line, so the boundary is the only place to check. For rules like these, continuity at the boundary means one thing: both rules give the same value there.

It does not matter which rule owns the boundary. Once both rules give 4 at x = 2, f(2) = 4 whichever inequality carries the "or equal to".

An unknown in the other rule

The unknown can be in either rule, and the method does not change. Take $f\left(x\right)=x^2+c$f(x) = x² + c for x < 1, and f(x) = 4x − 1 for $x\ge 1$x ≥ 1. At x = 1 the left rule gives $1^2+c=1+c$1² + c = 1 + c, and the right rule gives 4 × 1 − 1 = 3. Set them equal: 1 + c = 3, so c = 2. Check: $1^2+2=3$1² + 2 = 3.

The unknown can also multiply x. Take g(x) = ax − 2 for x < 2, and g(x) = x + 4 for $x\ge 2$x ≥ 2. At x = 2 the left rule gives 2a − 2 and the right rule gives 2 + 4 = 6. Set them equal: 2a − 2 = 6, so 2a = 8 and a = 4. Changing a does not slide the left-hand piece; it turns it about the point (0, −2), where ax − 2 = −2 for every a, until its end reaches (2, 6).

The gold graph is g with a = 4: the left piece rises from (0, −2) to meet the right piece at (2, 6). The white line is the left piece with a = 2, turned about the same point (0, −2), which reaches only (2, 2).

The usual mistakes

Taking k to be the height the pieces meet at. They meet at 4, but k is the constant in x + k, and 2 + k = 4 gives k = 2.

Taking k to be the boundary. The boundary 2 is the input where the rule changes. It is a coincidence of this example that k = 2 as well: for f(x) = 3x for x < 2 and x + k for $x\ge 2$x ≥ 2, the left rule gives 6, so 2 + k = 6 and k = 4.

Adding where the rule multiplies. The left rule is 2x, so at x = 2 it gives 2 × 2 = 4, not 2 + 2.