Piecewise Functions: AQA Level 2 Further Maths revision notes

Choose the rule from the input interval

A piecewise function uses different rules on different parts of its domain. Check which interval the input lies in before calculating the output. At a boundary value such as x=2, use the rule whose interval includes it: x\le2 includes 2, but x<2 does not.

Worked example 1

The function f is defined by

f(x)=\begin{cases}x+3 & -3\le x<0 \\ 4 & 0\le x<2 \\ 6-x & 2\le x\le5\end{cases}

Find f(-1), f(0) and f(2).

-1 belongs to the first interval, so f(-1)=-1+3=2
Your turn. What is f(0)? Use the rule whose interval includes 0.

Your answer

Zero belongs to 0\le x<2, so f(0)=4
Two belongs to the last interval, so f(2)=6-2=4
Answer: f(-1)=2, f(0)=4, f(2)=4

Explore which piece is active

An open circle marks a point that is not part of the graph; a filled circle marks one that is. Where one piece stops at an open circle and the next starts at a filled circle in the same place, as at (2,4), the graph simply carries on through that point: draw it filled.

Worked example 2

The function f is defined by

f(x)=\begin{cases}x+3 & -3\le x<0 \\ 4 & 0\le x<2 \\ 6-x & 2\le x\le5\end{cases}

Draw the graph of f, marking its endpoints correctly.

The first segment goes from filled (-3,0) to open (0,3). The second is horizontal at height 4, starting with filled (0,4)
The second piece stops short of x=2, so it ends with an open circle at (2,4). The last rule starts at x=2 with the same height, 6-2=4, so the point (2,4) is drawn filled and the graph is joined there. From (2,4) the last piece slopes down to x=5
Your turn. Calculate the final height, 6-5

Your answer

The final endpoint is filled (5,1)
Answer: the three separate segments described above. Do not join (0,3) to (0,4) with a vertical line: that would assign several outputs to one input.

Worked example 3

The function f is defined by

f(x)=\begin{cases}x+3 & -3\le x<0 \\ 4 & 0\le x<2 \\ 6-x & 2\le x\le5\end{cases}

State the domain and range of f.

The input intervals together cover every value from -3 to 5, including both endpoints.
The first piece produces 0\le f(x)<3. The middle piece gives 4. The last produces every output from 1 to 4, inclusive.
Your turn. What is the greatest output of the whole function?

Your answer

The greatest output is 4
Answer: domain -3\le x\le5; range 0\le f(x)\le4. There are no gaps in the combined set of outputs.

Solve on each piece, then check the interval

Solving a piece's formula can give a value of x outside the interval where that formula applies, and such a value is not a solution. Solve each piece that could give the required output, keep only the values of x that lie in that piece's own interval, and list the ones that remain.

Worked example 4

The function f is defined by

f(x)=\begin{cases}x+3 & -3\le x<0 \\ 4 & 0\le x<2 \\ 6-x & 2\le x\le5\end{cases}

Solve f(x)=2.

First piece: x+3=2 gives x=-1, which is in -3\le x<0. The constant middle piece has output 4, so gives no solution.
Your turn. Solve 6-x=2 on 2\le x\le5

Your answer

x=4, which lies in the last interval.
Answer: x=-1 or x=4. The horizontal line y=2 meets the graph twice.

Worked example 5

The function h is defined by

h(x)=\begin{cases}2x+1 & x<3 \\ k-x & x\ge3\end{cases}

Find k so the two pieces meet without a jump.

Your turn. What value does 2x+1 give when x=3? The left-hand piece never reaches x=3, but it gets as close as you like to this height.

Your answer

2(3)+1=7
The right-hand piece does include x=3, where its value is k-3. For the two pieces to meet with no jump, this must equal 7: k-3=7
Answer: k=10. Both formulas then give the height 7 at x=3, so the graph is joined there.

Worked example 6

The function g is defined by

g(x)=\begin{cases}x^2+1 & -2\le x<1 \\ 5-x & 1\le x\le4\end{cases}

Solve g(x)=3.

The quadratic piece gives x^2+1=3, so x=\pm\sqrt2. But \sqrt2>1, outside that piece.
Your turn. Which quadratic root lies in -2\le x<1?

Your answer

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Keep x=-\sqrt2
The linear piece gives 5-x=3, so x=2, which is in its interval.
Answer: x=-\sqrt2 or x=2. Each solution must pass its own interval check.

Common mistakes

Choosing a rule from the output. Use the input’s interval to choose the rule.
Filling an excluded endpoint. Use an open circle for a strict boundary such as x<2.
Joining a jump with a vertical line. Leave the gap; a vertical join would give one input several outputs.
Leaving overlapping range pieces uncombined. Combine the outputs from every piece into one set.
Keeping a solution outside its piece. Check each candidate against the interval of the rule used.