Composite and Inverse Functions

\mathrm{fg}(x) means apply g first, then apply f to the result.

1. Do g first: \mathrm{g}(x)=5x
2. Then f adds 8: \mathrm{f}(5x)=5x+8
3. (5(x+8)=5x+40 is \mathrm{gf}(x), the wrong order.)
4. Answer: \mathrm{fg}(x)=5x+8

\mathrm{gf}(x) means apply f first, then put the whole of \mathrm{f}(x) into g.

1. \mathrm{gf}(x)=\mathrm{g}(2x-5), so do f first.
2. \mathrm{g}(2x-5)=(2x-5)^2+1
3. (2x-5)^2=4x^2-20x+25
4. \mathrm{gf}(x)=4x^2-20x+26
5. (2x^2-3 is \mathrm{fg}(x), the wrong order; 4x^2+26 and 4x^2-10x+26 come from squaring the bracket wrongly.)

Work out \mathrm{f}(-3) first, then put your answer into g.

1. \mathrm{gf}(-3) means do f first
2. \mathrm{f}(-3)=(-3)^2-6=9-6=3
3. \mathrm{g}(3)=10-4\times3=10-12
4. Answer: -2

\mathrm{f}^{-1}(5) is the input that f turns into 5, so solve \mathrm{f}(x)=5.

1. \mathrm{f}^{-1}(5) is the value of x for which \mathrm{f}(x)=5
2. \dfrac{x+7}{3}=5
3. Multiply by 3: x+7=15
4. Answer: 8

Write y=4(2-x) and make x the subject, taking care with the minus sign in front of x.

1. Let y=4(2-x)
2. Divide by 4: \dfrac{y}{4}=2-x
3. Add x and subtract \dfrac{y}{4}: x=2-\dfrac{y}{4}
4. Write it in terms of x: \mathrm{f}^{-1}(x)=2-\dfrac{x}{4}
5. Answer: \mathrm{f}^{-1}(x)=\dfrac{8-x}{4}

Replace every x in \mathrm{f}(x) with (x+2), then expand and simplify.

1. \mathrm{fg}(x)=\mathrm{f}(x+2)
2. =(x+2)^2-4(x+2)
3. =x^2+4x+4-4x-8
4. Answer: \mathrm{fg}(x)=x^2-4

For \mathrm{f}(2x), replace x with (2x) in brackets, so the 2 is squared as well.

1. \mathrm{f}(2x)=3(2x)^2-5=12x^2-5
2. 2\mathrm{f}(x)=2(3x^2-5)=6x^2-10
3. 12x^2-5-(6x^2-10)=12x^2-5-6x^2+10
4. Answer: 6x^2+5

Write \mathrm{g}(x)=ax+b, so that \mathrm{gf}(x)=a(2x-1)+b, and compare this with 6x+4.

1. Let \mathrm{g}(x)=ax+b
2. \mathrm{gf}(x)=a(2x-1)+b=2ax-a+b
3. Compare the x terms with 6x+4: 2a=6, so a=3
4. Compare the numbers: -3+b=4, so b=7
5. Answer: \mathrm{g}(x)=3x+7

Write y=5+\sqrt{2x}, get the square root on its own, then square both sides.

1. Let y=5+\sqrt{2x}
2. Subtract 5: y-5=\sqrt{2x}
3. Square both sides: (y-5)^2=2x
4. Divide by 2: x=\dfrac{(y-5)^2}{2}
5. Answer: \mathrm{f}^{-1}(x)=\dfrac{(x-5)^2}{2}

The inputs of \mathrm{f}^{-1} are the outputs of f, so work out the smallest value of \mathrm{f}(x) when x\geqslant1.

1. The domain of \mathrm{f}^{-1} is the range of f
2. For x\geqslant1, 2x^2+3 gets bigger as x gets bigger
3. Smallest value: \mathrm{f}(1)=2+3=5
4. So the range of f is \mathrm{f}(x)\geqslant5
5. (x\geqslant1 is the domain of f, not of \mathrm{f}^{-1}; x\geqslant3 ignores the restriction x\geqslant1.)
6. Answer: x\geqslant5

Write y=\dfrac{4x+3}{5x-4}, multiply both sides by (5x-4) and collect the x terms on one side.

1. Let y=\dfrac{4x+3}{5x-4}
2. Multiply by (5x-4): 5xy-4y=4x+3
3. Collect the x terms: 5xy-4x=4y+3
4. Factorise: x(5y-4)=4y+3
5. Divide: x=\dfrac{4y+3}{5y-4}
6. So \mathrm{f}^{-1}(x)=\dfrac{4x+3}{5x-4}, which is the same as \mathrm{f}(x)
7. (\dfrac{5x-4}{4x+3} is \dfrac{1}{\mathrm{f}(x)}, which is not the inverse.)
8. Answer: \dfrac{4x+3}{5x-4}

Write \mathrm{gf}(a) in terms of a, set it equal to 7 and square both sides.

1. \mathrm{gf}(a)=\mathrm{g}(5a-2)=\sqrt{5a-2+6}
2. =\sqrt{5a+4}
3. \sqrt{5a+4}=7, so 5a+4=49
4. 5a=45
5. Answer: a=9

Write y=\dfrac{3x}{x+2}, multiply both sides by (x+2), then collect the x terms on one side and factorise.

1. Let y=\dfrac{3x}{x+2}
2. Multiply by (x+2): \;y(x+2)=3x
3. Expand: \;xy+2y=3x
4. Collect the x terms: \;2y=3x-xy
5. Factorise: \;2y=x(3-y)
6. Divide: \;x=\dfrac{2y}{3-y}
7. Swap back to x: \;\mathrm{f}^{-1}(x)=\dfrac{2x}{3-x}

Work out \mathrm{fg}(x) first, then set it equal to x-2 and rearrange into a quadratic equal to zero.

1. \mathrm{fg}(x)=(x-2)^2+3(x-2)
2. =x^2-4x+4+3x-6=x^2-x-2
3. Set equal: x^2-x-2=x-2
4. x^2-2x=0
5. Factorise: x(x-2)=0
6. (Don't divide by x: that loses the solution x=0.)
7. Answer: x=0 or x=2

Complete the square first, and use x\leqslant-2 to decide which square root to take.

1. Complete the square: x^2+4x-1=(x+2)^2-5
2. Let y=(x+2)^2-5, so (x+2)^2=y+5
3. x+2=\pm\sqrt{y+5}
4. x\leqslant-2 means x+2\leqslant0, so take the negative root
5. x=-2-\sqrt{y+5}
6. Answer: \mathrm{f}^{-1}(x)=-2-\sqrt{x+5}

\mathrm{fg}(x) means f applied to \mathrm{g}(x), so write \mathrm{f}(\mathrm{g}(x)) in terms of \mathrm{g}(x) and compare it with 6x^2-4. Answer part (a) before attempting part (b).

Part (a)

1. \mathrm{fg}(x)=\mathrm{f}(\mathrm{g}(x))=3\mathrm{g}(x)+2
2. So 3\mathrm{g}(x)+2=6x^2-4
3. 3\mathrm{g}(x)=6x^2-6
4. Answer: \mathrm{g}(x)=2x^2-2

Part (b)

1. \mathrm{gf}(x)=2(3x+2)^2-2
2. =2(9x^2+12x+4)-2=18x^2+24x+6
3. Set equal: 18x^2+24x+6=2x^2-2
4. 16x^2+24x+8=0
5. Divide by 8: 2x^2+3x+1=0
6. Factorise: (2x+1)(x+1)=0
7. Answer: x=-\dfrac12 or x=-1

Put \dfrac{6}{x+1} in place of x in f, then write the bottom of the big fraction as a single fraction. Answer part (a) before attempting part (b).

Part (a)

1. \mathrm{ff}(x)=\mathrm{f}\left(\dfrac{6}{x+1}\right)=\dfrac{6}{\frac{6}{x+1}+1}
2. Bottom: \dfrac{6}{x+1}+1=\dfrac{6+x+1}{x+1}=\dfrac{x+7}{x+1}
3. \mathrm{ff}(x)=6\div\dfrac{x+7}{x+1}=\dfrac{6(x+1)}{x+7}
4. Answer: \mathrm{ff}(x)=\dfrac{6x+6}{x+7}

Part (b)

1. \dfrac{6x+6}{x+7}=x
2. Multiply by (x+7): 6x+6=x^2+7x
3. x^2+x-6=0
4. Factorise: (x+3)(x-2)=0
5. Answer: x=-3 or x=2

Work out \mathrm{gf}(x) too, collect all the terms on one side of the inequality and factorise. Answer part (a) before attempting part (b).

Part (a)

1. \mathrm{fg}(x)=\mathrm{f}((x-2)^2)=3(x-2)^2+1
2. (x-2)^2=x^2-4x+4
3. 3x^2-12x+12+1
4. Answer: \mathrm{fg}(x)=3x^2-12x+13

Part (b)

1. \mathrm{gf}(x)=\mathrm{g}(3x+1)=(3x-1)^2=9x^2-6x+1
2. 3x^2-12x+13>9x^2-6x+1
3. 0>6x^2+6x-12
4. Divide by 6: x^2+x-2<0
5. Factorise: (x+2)(x-1)<0
6. The quadratic is negative between its roots -2 and 1
7. Answer: -2<x<1

Find \mathrm{f}^{-1}(x) and multiply it by 4 before you subtract, then take out the factor 4 and complete the square. Answer part (a) before attempting part (b).

Part (a)

1. \mathrm{gf}(x)=\mathrm{g}(2x+5)=(2x+5)^2-(2x+5)
2. (2x+5)^2=4x^2+20x+25
3. 4x^2+20x+25-2x-5
4. Answer: \mathrm{gf}(x)=4x^2+18x+20

Part (b)

1. y=2x+5 gives x=\dfrac{y-5}{2}, so \mathrm{f}^{-1}(x)=\dfrac{x-5}{2}
2. 4\mathrm{f}^{-1}(x)=2(x-5)=2x-10
3. \mathrm{gf}(x)-4\mathrm{f}^{-1}(x)=4x^2+18x+20-(2x-10)
4. =4x^2+16x+30
5. =4(x^2+4x)+30
6. =4\left((x+2)^2-4\right)+30
7. =4(x+2)^2-16+30
8. Answer: 4(x+2)^2+14

\mathrm{f}^{-1}(15)=4 tells you that \mathrm{f}(4)=15. For (b), do g first in \mathrm{fg}(x) and f first in \mathrm{gf}(x), and expand (2x+7)^2 carefully.

Part (a)

1. \mathrm{f}^{-1}(15)=4 means \mathrm{f}(4)=15
2. 2(4)+k=15
3. 8+k=15
4. k=7

Part (b)

1. With k=7: \;\mathrm{f}(x)=2x+7
2. \mathrm{fg}(x)=2(x^2+4)+7=2x^2+15
3. \mathrm{gf}(x)=(2x+7)^2+4=4x^2+28x+53
4. Set equal: \;2x^2+15=4x^2+28x+53
5. Rearrange: \;2x^2+28x+38=0, so x^2+14x+19=0
6. Quadratic formula: \;x=\dfrac{-14\pm\sqrt{14^2-4(1)(19)}}{2}=\dfrac{-14\pm\sqrt{120}}{2}
7. x=-1.52 or x=-12.48 (2 d.p.)