Formal, inverse and composite functions

(a) $7$

1. $f(4) = 3 \times 4 - 5 = 7$.

• B1 The correct answer, $7$.

(b) $7$

1. $fg(2)$ means apply $g$ first: $g(2) = 4$.
2. Then $f(4) = 7$.

• M1 Working out $g(2) = 4$ first.
• A1 The correct answer, $7$.

(c) $f^{-1}(x) = \dfrac{x + 5}{3}$

1. Write $y = 3x - 5$ and make $x$ the subject.
   $$y + 5 = 3x \Rightarrow x = \frac{y + 5}{3}$$
2. So $f^{-1}(x) = \frac{x + 5}{3}$.

• M1 Rearranging $y = 3x - 5$ to $x = \frac{y + 5}{3}$ (or reversing the operations).
• A1 $\frac{x + 5}{3}$.

(a) $\frac{2x + 7}{3}$

1. $$2y=3x-7$$
2. Write y = (3x $-$ 7)/2 and rearrange: 2y + 7 = 3x.
3. The original input is x = (2y + 7)/3.
4. Replace the input symbol y with x: $f^{-1}$(x) = (2x + 7)/3.

• P1 Establishing $2y=3x-7$ or an equivalent valid method.
• A1 Correct answer: $\frac{2x + 7}{3}$

(a) $4x^2 + 4x - 2$

1. $gf(x)$ means $g$ applied to $f(x)$.
   $$gf(x) = (2x + 1)^2 - 3$$
2. Expand.
   $$4x^2 + 4x + 1 - 3 = 4x^2 + 4x - 2$$

• M1 Writing $(2x + 1)^2 - 3$.
• A1 $4x^2 + 4x - 2$.

(b) $x = \pm\sqrt{6}$

1. $fg(x)$ means $f$ applied to $g(x)$.
   $$fg(x) = 2(x^2 - 3) + 1 = 2x^2 - 5$$
2. $$2x^2 - 5 = 7 \Rightarrow x^2 = 6 \Rightarrow x = \pm\sqrt{6}$$

• P1 Writing $fg(x) = 2(x^2 - 3) + 1$.
• P1 Forming and rearranging $2x^2 - 5 = 7$ to $x^2 = 6$.
• A1 $x = \sqrt{6}$ and $x = -\sqrt{6}$.

(a) $\frac{7}{3}$

1. $$(3x-1)^{2}=36$$
2. $$3x-1=6$$
3. The composite equation is (3x $-$ 1)$^{2}$ + 2 = 38, so (3x $-$ 1)$^{2}$ = 36.
4. The input to f must be non-negative, so 3x $-$ 1 $\ge$ 0. Therefore 3x $-$ 1 = 6, not $-6.$
5. Solve 3x = 7 to obtain x = 7/3.

• P1 Establishing $(3x-1)^{2}=36$ or an equivalent valid method.
• P1 Establishing $3x-1=6$ or an equivalent valid method.
• A1 Correct answer: $\frac{7}{3}$

(a) $\frac{x+5}{3}$

1. Reverse subtraction, then multiplication.
   $$y=(x+5)/3$$
2. Therefore $\frac{x+5}{3}$.

• M1 Reverse subtraction, then multiplication.
• A1 Correct answer: $\frac{x+5}{3}$

(b) $-3,3$

1. The one-to-one linear function can be undone on both sides.
   $$x^{2}+1=10$$
2. Keep both square roots.
   $$x^{2}=9$$
3. Therefore $-3,3$.

• M1 The one-to-one linear function can be undone on both sides.
• M1 Keep both square roots.
• A1 Correct answer: $-3,3$

(a) $f^{-1}(x)=\sqrt{x-6}$ for $x\ge 6$.

1. Rearrange the quadratic output relation and retain the non-negative input branch.
   $$y=\sqrt{x-6}$$
2. Therefore $f^{-1}(x)=\sqrt{x-6}$ for $x\ge 6$.

• M1 Rearrange the quadratic output relation and retain the non-negative input branch.
• A1 Correct answer: $f^{-1}(x)=\sqrt{x-6}$ for $x\ge 6$.

(b) $\frac{5}{4}$

1. The input g(x) to f must be non-negative, so use the positive square root.
   $$4x-1=4$$
2. Therefore $\frac{5}{4}$.

• M1 The input g(x) to f must be non-negative, so use the positive square root.
• A1 Correct answer: $\frac{5}{4}$

(a) $f^{-1}(x) = \dfrac{x + 4}{x - 1}$

1. Let $y = \frac{x + 4}{x - 1}$ and make $x$ the subject.
   $$y(x - 1) = x + 4 \Rightarrow xy - y = x + 4$$
2. Collect the $x$ terms.
   $$xy - x = y + 4 \Rightarrow x(y - 1) = y + 4 \Rightarrow x = \frac{y + 4}{y - 1}$$
3. So $f^{-1}(x) = \frac{x + 4}{x - 1}$.

• M1 Multiplying out: $xy - y = x + 4$.
• M1 Collecting and factorising: $x(y - 1) = y + 4$.
• A1 $f^{-1}(x) = \frac{x + 4}{x - 1}$.

(b) $f$ is its own inverse, so $ff(x) = x$.

1. $f^{-1} = f$, so applying $f$ twice undoes itself: $ff(x) = f^{-1}f(x) = x$.

• C1 Saying $f$ is its own inverse, so $ff(x) = x$.

(a) $a = \frac{5}{3}$

1. Find the inverse.
   $$y = 5 - 2x \Rightarrow x = \frac{5 - y}{2} \Rightarrow f^{-1}(x) = \frac{5 - x}{2}$$
2. Set them equal.
   $$5 - 2a = \frac{5 - a}{2} \Rightarrow 10 - 4a = 5 - a$$
3. $$5 = 3a \Rightarrow a = \frac{5}{3}$$

• P1 Finding $f^{-1}(x) = \frac{5 - x}{2}$.
• P1 Forming $5 - 2a = \frac{5 - a}{2}$.
• P1 Clearing the fraction and collecting: $3a = 5$.
• A1 $a = \frac{5}{3}$.