Composite Functions

Worked example

1. 1

   $f(x)=x^2-4,g(x)=2x+1$ — the two functions; find fg(3)

2. 2

   $g(3)=2(3)+1=7$ — g first: it is next to the 3

3. 3

   $f(7)=7^2-4=45$ — then f takes the 7

Worked example

1. 1

   $fg(x)=f(2x+1)$ — g(x) goes into f, so the x in f is replaced by g(x).

2. 2

   $fg(x)=(2x+1{)}^2-4$ — Every x in f becomes (2x + 1). Keep the bracket.

3. 3

   $fg(x)=(2x+1)(2x+1)-4$ — A square means times itself, so write the bracket twice.

4. 4

   $fg(x)=4x^2+2x+2x+1-4$ — Multiply each term of one bracket by each term of the other.

5. 5

   $fg(x)=4x^2+4x-3$ — Collect like terms: 2x + 2x = 4x, and 1 − 4 = −3.

Worked example

1. 1

   $p(x)=4x-7;\text{find }pp(-1)$ — pp(−1) means p(p(−1)). Work out the inside p(−1) first.

2. 2

   $p(-1)=4(-1)-7$ — Put −1 into p. Use a bracket to keep the negative sign.

3. 3

   $p(-1)=-4-7=-11$ — 4 × −1 = −4, then −4 − 7 = −11.

4. 4

   $p(-11)=4(-11)-7$ — Now put that answer, −11, into p.

5. 5

   $p(-11)=-44-7=-51$ — 4 × −11 = −44, then −44 − 7 = −51.

Worked example

1. 1

   $pp(x)=p(4x-7)$ — For the expression, put the whole of p(x) into p.

2. 2

   $pp(x)=4(4x-7)-7$ — Every x in p becomes (4x − 7). Keep the bracket.

3. 3

   $pp(x)=16x-28-7$ — Multiply out the bracket: 4 × 4x and 4 × −7.

4. 4

   $pp(x)=16x-35$ — −28 − 7 = −35.

Worked example

1. 1

   $4x^2+4x-3=5$ — This is fg(x) from the fg(x) card, set equal to 5.

2. 2

   $4x^2+4x-8=0$ — Take 5 from both sides, so one side is 0.

3. 3

   $x^2+x-2=0$ — Divide every term by 4 to make the numbers smaller.

4. 4

   $(x+2)(x-1)=0$ — Factorise: the numbers +2 and −1 multiply to −2 and add to 1.

5. 5

   $x=-2\text{or}x=1$ — One of the brackets must be 0, so solve each one.

6. 6

   $g(-2)=2(-2)+1=-3,f(-3)=(-3{)}^2-4=5$ — Check x = −2: g first, then f. The answer is 5, so it works.

7. 7

   $g(1)=2(1)+1=3,f(3)=3^2-4=5$ — Check x = 1 the same way. It also gives 5.

Worked example

1. 1

   $4x^2+4x-3>0$ — This is fg(x) > 0. First find where it equals 0.

2. 2

   $(2x+3)(2x-1)=0$ — Factorise the expression.

3. 3

   $x=-\frac{3}{2}\text{or}x=\frac{1}{2}$ — Set each bracket to 0: 2x + 3 = 0 gives −3/2, and 2x − 1 = 0 gives 1/2. These are the crossing points.

4. 4

   $4x^2\text{has a positive number in front}$ — So the graph is a U-shape, opening upwards.

5. 5

   $x<-\frac{3}{2}\text{or}x>\frac{1}{2}$ — A U-shape is above 0 outside the two crossing points.

Worked example

1. 1

   $f(x)=x^2-2x,g(x)=x+3$ — Here h = fg for x ≥ −2. First put g into f.

2. 2

   $h(x)=(x+3{)}^2-2(x+3)$ — Every x in f becomes (x + 3). Keep the brackets.

3. 3

   $h(x)=x^2+6x+9-2x-6$ — Expand (x + 3)². Then multiply both terms of the second bracket by −2.

4. 4

   $h(x)=x^2+4x+3$ — Collect like terms.

5. 5

   $h(x)=(x+2{)}^2-4+3$ — Complete the square: half of 4 is 2, and (x + 2)² has an extra 4, so take 4 away.

6. 6

   $h(x)=(x+2{)}^2-1$ — −4 + 3 = −1.

7. 7

   $y+1=(x+2{)}^2$ — Write y for h(x), then add 1 to both sides.

8. 8

   $\pm \sqrt{y+1}=x+2$ — Square root both sides. A square has two roots.

9. 9

   $x=-2+\sqrt{y+1}$ — x ≥ −2 means x + 2 is 0 or more, so keep the plus. Then take 2 from both sides.

10. 10

    $h^{-1}(x)=-2+\sqrt{x+1}$ — Swap y for x.

Worked example

1. 1

   $f(x)=2x+1\Rightarrow f^{-1}(x)=\frac{x-1}{2}$ — Undo +1 with −1, then undo ×2 with ÷2.

2. 2

   $g(x)=x-3\Rightarrow g^{-1}(x)=x+3$ — Undo −3 with +3.

3. 3

   $(fg{)}^{-1}(x)=g^{-1}\left(f^{-1}(x)\right)$ — f⁻¹ goes in first, because f was the last function to be applied.

4. 4

   $(fg{)}^{-1}(x)=\frac{x-1}{2}+3$ — Put f⁻¹(x) into g⁻¹, which adds 3.

5. 5

   $(fg{)}^{-1}(x)=\frac{x-1}{2}+\frac{6}{2}$ — Write 3 as 6/2, so the fractions have the same bottom.

6. 6

   $(fg{)}^{-1}(x)=\frac{x+5}{2}$ — Add the tops: x − 1 + 6 = x + 5.

7. 7

   $fg(x)=2(x-3)+1=2x-5\Rightarrow \frac{x+5}{2}$ — Check: expand fg first, then swap and solve 2x − 5. It gives the same answer.