Functions

(a) $6 \times 8 = 48$, then $48 - 5 = 43$

(b) After the first box: $17 + 13 = 30$

The second box must turn 30 into 10: $30 \div 3 = 10$ (or $30 - 20 = 10$)

So the missing box is $\div 3$ (or $-20$).

(a) $11 + 5 = 16$, then $16 \div 2 = 8$

(b) Work backwards from the output: $28 \div 4 = 7$

The first box turns 10 into 7, so it is $-3$ (or $\times 0.7$).

(a) $28 \div 7 = 4$, then $4 + 5 = 9$.

(b) Work backwards from 154: $154 \div 11 = 14$.

The missing step turns 8 into 14, so it is $+6$.

(a) $7 \times 2 = 14$, then $14 - 3 = 11$.

(b) Work backwards with inverses: $41 + 3 = 44$, then $44 \div 2 = 22$.

(a) $8 \times 5 = 40$, $40 - 2 = 38$

(b) Work backwards: $28 + 2 = 30$, $30 \div 5 = 6$

(a) $13 \times 2 = 26$, $26 - 10 = 16$

(b) Work backwards: $28 + 10 = 38$, $38 \div 2 = 19$

(c) Solve $2x - 10 = x$ to get $x = 10$. Check: $10 \times 2 = 20$, $20 - 10 = 10$.

(a) $4 \times 70 + 50 = 280 + 50 = £330$

(b) Undo the rule: $680 - 50 = 630$, then $630 \div 70 = 9$ weeks.

$f(x) = \frac{2}{x^2}$ and $g(x) = 4x^3$

(a) $f(-5) = \frac{2}{(-5)^2} = \frac{2}{25}$

(b) $g(1) = 4$, then $f(4) = \frac{2}{16} = \frac{1}{8}$

(a) Let $y = \frac{5x - 3}{4}$. Then $4y = 5x - 3$, $5x = 4y + 3$, $x = \frac{4y + 3}{5}$

So $f^{-1}(x) = \frac{4x + 3}{5}$

(b) $h(5) = 1 - 10 = -9$

$g(-9) = (-9 - 1)^2 = 100$

$f(x) = \dfrac{12}{x+1}$, $g(x) = 5 - 3x$

(a) $f(-3) = \dfrac{12}{-2} = -6$

(b) $g(1) = 2$, so $fg(1) = f(2) = \dfrac{12}{3} = 4$

(c) Solve $5 - 3x = 4$: $x = \frac13$

(a) Let $y = \frac{\sqrt{x} + 2}{5}$. Then $5y = \sqrt{x} + 2$, so $\sqrt{x} = 5y - 2$ and $x = (5y - 2)^2$

$g^{-1}(x) = (5x - 2)^2$

(b) $gf(x) = 3$ means $f(x) = g^{-1}(3) = (15 - 2)^2 = 169$

$3x + 4 = 169$, so $3x = 165$ and $x = 55$

(a) $f(-3) = (2(-3) + 3)^2 = (-3)^2 = 9$, then $g(9) = 2 \times 9 - 1 = 17$.

(b) Let $y = 2x - 1$. Then $y + 1 = 2x$, so $x = \frac{y + 1}{2}$.

$g^{-1}(x) = \frac{x + 1}{2}$

(a) $g(16) = \sqrt[3]{32 - 5} = \sqrt[3]{27} = 3$

(b) Let $y = \sqrt[3]{2x - 5}$. Cube: $y^3 = 2x - 5$, so $x = \frac{y^3 + 5}{2}$

$g^{-1}(x) = \frac{x^3 + 5}{2}$

$hg^{-1}(x) = \frac{1}{\frac{x^3 + 5}{2}} = \frac{2}{x^3 + 5}$

(a) $g(5) = 3(2 \times 5 + 1) = 3 \times 11 = 33$

(b) $f(9) = \frac{12}{\sqrt{9}} = 4$, then $g(4) = 3(9) = 27$

(c) Solve $3(2x + 1) = 6$: $2x + 1 = 2$, so $x = \frac{1}{2}$

(a) $y = 3x - 1 \Rightarrow x = \frac{y + 1}{3}$, so $f^{-1}(x) = \frac{x + 1}{3}$

(b) $fg(x) = 3(x^2 + 4) - 1 = 3x^2 + 11$

$gf(x) = (3x - 1)^2 + 4 = 9x^2 - 6x + 5$

$3x^2 + 11 = 2(9x^2 - 6x + 5) = 18x^2 - 12x + 10$

$0 = 15x^2 - 12x - 1$

(a) $f(3) = 2 \times 3^3 - 4 = 54 - 4 = 50$, so $f^{-1}(50) = 3$

(b) $hg(x) = (x+2)^2 = x^2 + 4x + 4$

$x^2 + 4x + 4 = 3x^2 + x - 1$

$0 = 2x^2 - 3x - 5 = (2x - 5)(x + 1)$

$x = 2.5$ or $x = -1$

(a) $gf(x) = g\big((x+1)^2\big) = 2\big((x+1)^2 - 1\big)$

$= 2(x^2 + 2x + 1 - 1) = 2(x^2 + 2x) = 2x(x + 2)$

(b) Solve $g(x) = 7$: $2(x - 1) = 7$, so $x - 1 = 3.5$ and $x = 4.5$

(a) $f(23) = 4\sin 23° = 1.56$

(b) $g(34) = 2 \times 34 - 3 = 65$, then $f(65) = 4\sin 65° = 3.63$

(c) $25$ has two square roots, $5$ and $-5$. So $x + 4 = -5$ is also possible, giving a second answer $x = -9$.

$f(x + 2) = 3(x + 2)^2 - 2(x + 2) - 8$

$= 3(x^2 + 4x + 4) - 2x - 4 - 8$

$= 3x^2 + 12x + 12 - 2x - 12$

$= 3x^2 + 10x$

$h(x) = f(g(x)) = \sqrt[3]{2x + 3}$

Let $y = \sqrt[3]{2x + 3}$, so $y^3 = 2x + 3$ and $x = \frac{y^3 - 3}{2}$

$h^{-1}(x) = \frac{x^3 - 3}{2}$

(a) $f(1) = 3 + 1 = 4$, then $g(4) = \frac{4}{16} = \frac{1}{4}$

(b) $fg(x) = 3\left(\frac{4}{x^2}\right)^2 + 1 = \frac{48}{x^4} + 1$

Let $y = \frac{48}{x^4} + 1$, so $x^4 = \frac{48}{y - 1}$

$h(x) = \sqrt[4]{\frac{48}{x - 1}}$