Exam practice

If f:A → B and g:B → C are functions, the function from A to C that takes each x in A to g(f(x)) is the composite function of f and g, written g ∘ f (or gf). It is f first, then g.
Example: f(x) = 2x + 1 from A = {1, 2, 3} to B = {3, 5, 7}, and $g(x) = x^2$ from B to C = {9, 25, 49}. Then $g \circ f(x) = (2x + 1)^2$ and g ∘ f = {(1, 9), (2, 25), (3, 49)}.

• a.g = {(1, 2), (2, 3), (3, 4), (4, 5)}
• b.f = {(3, −1), (4, 0), (2, −1), (5, 1)}
• c.f ∘ g(1) = f(2) = −1, f ∘ g(2) = f(3) = −1, f ∘ g(3) = f(4) = 0, f ∘ g(4) = f(5) = 1.
• d.f ∘ g = {(1, −1), (2, −1), (3, 0), (4, 1)}. (It is many-to-one: 1 and 2 both end at −1, and nothing ends at 2 in C.)

f first: 1 → 2 → 4, 2 → 3 → 5, 3 → 4 → 6. So gf = {(1, 4), (2, 5), (3, 6)}.

• a.$gf(x) = g(3x + 2) = 5 - 3(3x + 2) = 5 - 9x - 6 = \mathbf{-1 - 9x}$
• b.$fg(x) = f(5 - 3x) = 3(5 - 3x) + 2 = \mathbf{17 - 9x}$
• c.$ff(x) = f(3x + 2) = 3(3x + 2) + 2 = \mathbf{9x + 8}$
• d.$ff(3) = 9 \times 3 + 8 = \mathbf{35}$ (check: f(3) = 11, f(11) = 35 ✓)

Put the whole fraction g(x) in place of x in f, then multiply the top and the bottom by x − 4.

$$fg(x) = f\left(\frac{3x + 2}{x - 4}\right) = \frac{2 \cdot \frac{3x + 2}{x - 4} - 1}{\frac{3x + 2}{x - 4} + 3}$$
Multiply the top and the bottom by x − 4:
$$fg(x) = \frac{2(3x + 2) - (x - 4)}{(3x + 2) + 3(x - 4)} = \frac{6x + 4 - x + 4}{3x + 2 + 3x - 12} = \mathbf{\frac{5x + 8}{6x - 10}}$$
(for x ≠ 4 and $x \neq \frac{5}{3}$: at $x = \frac{5}{3}$, g(x) = −3, which f can’t take).

$$fg(x) = f\left(\frac{x - 5}{2}\right) = 2 \cdot \frac{x - 5}{2} + 5 = x - 5 + 5 = x$$
fg(x) = x gives every input back unchanged, so fg is the identity function. (g undoes f: it is the inverse of f, as 1.3 explains.)

• a.$fg(x) = f(g(x)) = \frac{g(x) - 3}{2} = x$, so g(x) − 3 = 2x and g(x) = 2x + 3. Check: $f(2x + 3) = \frac{2x}{2} = x$ ✓
• b.gh(x) = g(h(x)) = 4 − h(x) = 11 − 2x, so h(x) = 2x − 7. Check: g(2x − 7) = 4 − 2x + 7 = 11 − 2x ✓

• a.$f(5) = \frac{6}{3} = 2$, so gf(5) = g(2) = 4a − 1 = 7, and a = 2.
• b.f(5) = 5a + 5, so gf(5) = 8(5a + 5) + 13 = 40a + 53 = 93. Then 40a = 40 and a = 1.

f(1) = k + 1, so gf(1) = 3(k + 1) − 7 = 3k − 4 = 2. Then 3k = 6 and k = 2. Check: f(1) = 3, g(3) = 2 ✓

First find g(x) from fg(x), as in question 8.

fg(x) = 3g(x) − 2 = 6x − 2, so g(x) = 2x.
Then gf(x) = g(3x − 2) = 2(3x − 2) = 6x − 4 = 8, so 6x = 12 and x = 2. Check: f(2) = 4, g(4) = 8 ✓

If f:A → B is a one-to-one and onto function, the function from B to A that takes each image back to its pre-image is the inverse function of f, written $f^{-1}$: if f(x) = y, then $f^{-1}(y) = x$. Its mapping diagram is that of f with every arrow turned round.
Example: f(x) = x + 3 from A = {1, 2, 3} to B = {4, 5, 6}; then $f^{-1}(x) = x - 3$, from B to A.

$f^{-1} = \{(2, 1), (4, 2), (6, 3)\}$.

Yes. One-to-one: 2a − 3 = 2b − 3 gives a = b. Onto: any real y is $f\left(\frac{y + 3}{2}\right)$.
y = 2x − 3; swap: x = 2y − 3, so $f^{-1}(x) = \frac{x + 3}{2}$.

c. To find $h^{-1}$, swap, multiply out, and gather the y terms on one side. d. After finding $f^{-1}$, set the two fractions equal and cross-multiply; don’t lose a root by dividing by x.

• a.$f^{-1}(x) = \frac{x + 2}{5} = 2$, so x + 2 = 10 and x = 8. (Quicker: $f^{-1}(x) = 2$ means x = f(2) = 8.)
• b.$g^{-1}(x) = \frac{4x + 7}{3} = 5$, so 4x = 8 and x = 2. (Or $x = g(5) = \frac{8}{4} = 2$.)
• c.Swap: $x = \frac{2y - 3}{3y - 5}$, so 3xy − 5x = 2y − 3, y(3x − 2) = 5x − 3 and $h^{-1}(x) = \frac{5x - 3}{3x - 2}$.

Then $\frac{5x - 3}{3x - 2} = 2$: 5x − 3 = 6x − 4, so x = 1. (Or $x = h(2) = \frac{1}{1} = 1$.)
d. Swap: $x = \frac{y}{3 + 2y}$, so 3x + 2xy = y, y(1 − 2x) = 3x and $f^{-1}(x) = \frac{3x}{1 - 2x}$.
Then $\frac{3x}{1 - 2x} = \frac{x}{3 + 2x}$: 3x(3 + 2x) = x(1 − 2x), $9x + 6x^2 = x - 2x^2$, $8x^2 + 8x = 0$, 8x(x + 1) = 0.
So x = 0 or x = −1. Check: $f(-1) = \frac{-1}{1} = -1$ and $f^{-1}(-1) = \frac{-3}{3} = -1$ ✓

In b, the function written next to the bracket works first: $g^{-1}(2)$, then $f^{-1}$ of that. In c, find gh(x) first, simplify it, then find its inverse.

• a.$g^{-1}(x) = \frac{5x + 3}{2}$. So $g^{-1}f(x) = g^{-1}(7x - 1) = \frac{5(7x - 1) + 3}{2} = \mathbf{\frac{35x - 2}{2}}$.
• b.$g^{-1}(x) = 5x - 2$, so $g^{-1}(2) = 8$. $f^{-1}(x) = \frac{x + 3}{4}$, so $f^{-1}(8) = \mathbf{\frac{11}{4}}$.
• c.$gh(x) = \frac{3h - 1}{h - 1}$ with $h = \frac{2x - 1}{x - 3}$; multiplying top and bottom by x − 3: $gh(x) = \frac{3(2x - 1) - (x - 3)}{(2x - 1) - (x - 3)} = \frac{5x}{x + 2}$.

Its inverse: $x = \frac{5y}{y + 2}$, xy + 2x = 5y, y(5 − x) = 2x, so $(gh)^{-1}(x) = \frac{2x}{5 - x}$.
$hg(x) = \frac{2g - 1}{g - 3} = \frac{2(3x - 1) - (x - 1)}{(3x - 1) - 3(x - 1)} = \frac{5x - 1}{2}$.
Its inverse: $x = \frac{5y - 1}{2}$, so $(hg)^{-1}(x) = \frac{2x + 1}{5}$.
(The careful box’s rule checks out: $h^{-1}(x) = \frac{3x - 1}{x - 2}$ and $g^{-1}(x) = \frac{x - 1}{x - 3}$, and $h^{-1}(g^{-1}(x))$ also simplifies to $\frac{2x}{5 - x}$.)

• a.This is Example 5: ff(x) = 4x − 9 and $g^{-1}(x) = \frac{3x + 7}{2}$, so 8x − 18 = 3x + 7 and x = 5.
• b.ff(x) = 2(2x − 5) − 5 = 4x − 15; $g^{-1}(x) = \frac{2x - 5}{3}$. Then 3(4x − 15) = 2x − 5, 12x − 45 = 2x − 5, 10x = 40, so x = 4. Check: ff(4) = 1 and $g^{-1}(4) = 1$ ✓
• c.$g^{-1}(x) = 2x + 5$, so $fg^{-1}(x) = f(2x + 5) = \frac{2x + 6}{2} = x + 3 = 6$, and x = 3.