Piecewise Functions

Decide which part of the domain contains x=5 before you substitute.

1. 5>3, so use the second expression, 14-2x
2. \mathrm{f}(5)=14-2\times5
3. Answer: 4

You don't know k, so use the other part of f to find the y-coordinate at x=4.

1. x=4 is in the domain 0\leqslant x\leqslant 4 of the first part
2. \mathrm{f}(4)=9-\tfrac12\times4=9-2=7
3. Answer: P=(4,\ 7)

The domain is every value of x for which f is defined, so put the three parts of the domain together.

1. The three parts cover -3\leqslant x<0, then 0\leqslant x<2, then 2\leqslant x\leqslant 5
2. Together they run from -3 to 5 with no gaps, and both ends are included
3. Answer: -3\leqslant x\leqslant 5

Check the values x=0, x=2 and x=6: each must be in exactly one of the two parts.

1. Every x from 0 to 6 must be in exactly one part
2. 0\leqslant x\leqslant 2 and 2\leqslant x\leqslant 6 both contain x=2, giving two values (3 and 5)
3. 0\leqslant x<2 and 2<x\leqslant 6 leave out x=2
4. 0\leqslant x<2 and 2\leqslant x<6 leave out x=6
5. Answer: 0\leqslant x<2 and 2\leqslant x\leqslant 6

At a boundary, look carefully at the inequality signs to decide which piece includes that value of x.

1. x=1 satisfies 1\leqslant x<3, so use x^2+3: \mathrm{f}(1)=1+3=4
2. x=3 satisfies x\geqslant3, so use 15-2x: \mathrm{f}(3)=15-6=9
3. \mathrm{f}(1)+\mathrm{f}(3)=4+9=13

Work out \mathrm{f}(-3), \mathrm{f}(0) and \mathrm{f}(3) and compare them with each graph.

1. x+3 rises from (-3,\ 0) to (-1,\ 2)
2. x^2+1 is a U-shaped curve with its lowest point at (0,\ 1)
3. 3-x falls from (1,\ 2) to (3,\ 0)
4. A has a \cap-shaped middle, B falls on the left and D rises on the right
5. Answer: Graph C

\mathrm{ff}(7) means f of \mathrm{f}(7): read \mathrm{f}(7) from the graph first.

1. From the graph, \mathrm{f}(7)=5
2. So \mathrm{ff}(7)=\mathrm{f}(5)
3. From the graph, \mathrm{f}(5)=6
4. Answer: 6

Solve \mathrm{f}(x)=1 for each part separately, then check each answer is in that part's domain.

1. First part: 2x+3=1 gives x=-1, and -1<1 ✓
2. Second part: 7-2x=1 gives x=3, and 3\geqslant1 ✓
3. Answer: x=-1 or x=3

Read two points on that part of the graph and work out its gradient first.

1. The part goes from (4,\ 4) to (8,\ 1)
2. Gradient =\dfrac{1-4}{8-4}=-\dfrac34
3. y-4=-\tfrac34(x-4), so y=-\tfrac34x+3+4
4. Answer: \mathrm{f}(x)=7-\tfrac34x

The pieces join, so both expressions give the same value when x=2

1. Substitute x=2 into both pieces.
2. First piece: 4a-3
3. Second piece: 6+a
4. They join, so 4a-3=6+a
5. 3a=9
6. a=3

Solve the inequality for each part, keep only the values inside that part's domain, then join the two sets of values together.

1. First part: 5-2x\leqslant3 gives x\geqslant1
2. With x<2 this gives 1\leqslant x<2
3. Second part: 2x-3\leqslant3 gives x\leqslant3
4. With x\geqslant2 this gives 2\leqslant x\leqslant3
5. Together: 1\leqslant x<2 and 2\leqslant x\leqslant 3
6. Answer: 1\leqslant x\leqslant 3

The distance travelled is the area under the speed–time graph: sketch the graph and split the area into simple shapes.

1. The graph rises from (0,\ 0) to (4,\ 8), is flat to (10,\ 8), then falls to (18,\ 0)
2. Triangle: \tfrac12\times4\times8=16
3. Rectangle: 6\times8=48
4. Triangle: \tfrac12\times8\times8=32
5. 16+48+32=96
6. Answer: 96 m

Substitute x=a into both expressions and set them equal: this gives a quadratic equation in a.

1. The parts join where x=a, so a+6=a^2
2. a^2-a-6=0
3. (a-3)(a+2)=0, so a=3 or a=-2
4. a is positive, so reject a=-2
5. Answer: a=3

Solve \mathrm{f}(x)=8 for each part and reject any solution that is not in that part's domain.

1. First part: x^2-2x=8, so x^2-2x-8=0
2. (x-4)(x+2)=0, so x=4 or x=-2
3. This part needs x\leqslant3: keep x=-2, reject x=4
4. Second part: 12-3x=8 gives x=\tfrac43
5. This part needs x>3, so reject x=\tfrac43
6. Answer: x=-2 only

The lowest point of the quadratic part is not at an end of its domain: complete the square to find it.

1. x^2+2x-3=(x+1)^2-4, so its minimum is -4, at x=-1 (inside -3\leqslant x\leqslant1)
2. At the ends: \mathrm{f}(-3)=0 and \mathrm{f}(1)=0, so this part gives -4\leqslant \mathrm{f}(x)\leqslant0
3. 2x-2 increases from 0 (not included) to \mathrm{f}(4)=6
4. So this part gives 0<\mathrm{f}(x)\leqslant6
5. Smallest value -4, largest value 6
6. Answer: -4\leqslant \mathrm{f}(x)\leqslant 6

Solve each part separately and keep only the solutions that lie in that part's domain; the graph shows how many to expect.

Part (a)

1. First part: 2x+8=3 gives x=-2.5, and -2.5<-2 ✓
2. Second part: x^2=3 gives x=\pm\sqrt3
3. \sqrt3=1.73\ldots, so both \pm\sqrt3 are in -2\leqslant x\leqslant2 ✓
4. Third part: 8-2x=3 gives x=2.5, and 2.5>2 ✓
5. Answer: x=-2.5,\ -\sqrt3,\ \sqrt3,\ 2.5

Part (b)

1. A horizontal line y=k meets the graph three times only when it touches the bottom of the curve
2. The lowest point of x^2 is (0,\ 0)
3. \mathrm{f}(x)=0 gives x=-4, x=0 and x=4
4. Answer: k=0

Part (c)

1. First part: 2x+8=4 gives x=-2, but this part needs x<-2, so reject
2. Second part: x^2=4 gives x=-2 and x=2 ✓
3. Third part: 8-2x=4 gives x=2, but this part needs x>2, so reject
4. Answer: 2 solutions

Set each part of f equal to 6-4x in turn, and reject any solution that is not in that part's domain.

Part (a)

1. First part: 2x+4=6-4x gives x=\tfrac13, but this part needs x<-1, so reject
2. Second part: 3-x^2=6-4x, so x^2-4x+3=0
3. (x-1)(x-3)=0, so x=1 or x=3
4. This part needs -1\leqslant x\leqslant2: keep x=1, reject x=3
5. At x=1: y=6-4=2
6. Answer: (1,\ 2)

Part (b)

1. Third part: 11-6x=6-4x
2. 5=2x, so x=2.5, and 2.5>2 ✓
3. At x=2.5: y=6-4\times2.5=-4
4. Answer: (2.5,\ -4)

Use the second part to find the y-coordinate where the parts join, then substitute both known points into ax^2+bx. Answer part (a) before attempting part (b).

Part (a)

1. At x=4 the second part gives 20-2\times4=12, so the parts join at (4,\ 12)
2. (4,\ 12): 16a+4b=12, so 4a+b=3
3. (2,\ 10): 4a+2b=10, so 2a+b=5
4. Subtract: 2a=-2
5. Answer: a=-1

Part (b)

1. Substitute a=-1 into 2a+b=5
2. -2+b=5
3. Answer: b=7

Part (c)

1. 7x-x^2=\tfrac{49}{4}-(x-\tfrac72)^2, so the curve has its maximum \tfrac{49}{4}=12.25 at x=3.5
2. x=3.5 is inside 0\leqslant x\leqslant 4, so the highest value is 12.25, not 12
3. The lowest values are \mathrm{f}(0)=0 and \mathrm{f}(10)=0
4. Answer: 0\leqslant \mathrm{f}(x)\leqslant \dfrac{49}{4}

At a turning point the gradient is zero, so differentiate the first part. Answer part (a) before attempting part (b).

Part (a)

1. For the first part, \dfrac{\mathrm{d}y}{\mathrm{d}x}=2x+a
2. At the turning point the gradient is 0: 2\times3+a=0
3. Answer: a=-6

Part (b)

1. At x=5 the second part gives 3\times5-11=4
2. First part at x=5: 25-30+b=4
3. Answer: b=9

Part (c)

1. \mathrm{f}(1)=1-6+9=4, so the point is (1,\ 4)
2. Gradient: 2\times1-6=-4
3. y-4=-4(x-1)
4. Answer: y=-4x+8

Use each linear piece to find the y-coordinate of the joining point, then substitute both joining points into x^2+px+q

Part (a)

1. At x=1 the first piece gives 4-1=3, so 1+p+q=3
2. At x=5 the third piece gives 2\times5-15=-5, so 25+5p+q=-5
3. Subtract the first equation from the second: 24+4p=-8
4. 4p=-32, so p=-8

Part (b)

1. Substitute p=-8 into 1+p+q=3
2. 1-8+q=3
3. q=10

Part (c)

1. 4-x=-1 gives x=5, but this piece needs x<1, so reject.
2. x^2-8x+10=-1 gives x^2-8x+11=0
3. x=\dfrac{8\pm\sqrt{64-44}}{2}=\dfrac{8\pm\sqrt{20}}{2}=4\pm\sqrt5
4. 4-\sqrt5=1.763\ldots is in 1\leqslant x\leqslant5 ✓; 4+\sqrt5=6.236\ldots is not, so reject.
5. 2x-15=-1 gives x=7, which satisfies x>5 ✓
6. x=1.76 or x=7