Exercise (Inverse of a function):

Exercise (Solution):

1. Find the inverse function of the following functions.

Question 1(a)

The given function is,

$$f = \{(1, 8), (2, 9), (3, 10)\}$$

$\textcolor{red}{[\text {The position of antecedent and consequent will be interchanged.]}}$

$$\therefore f^{-1} = \{(8, 1), (9, 2), (10, 3)\}$$

Question 1(b)

The given function is,

$$g = \{(8, 5), (9, 6), (10, 7)\}$$

$$\therefore g^{-1} = \{(5, 8), (6, 9), (7, 10)\}$$

Question 1(c)

The given function is,

$$h = \{(2, 5), (3, 6), (4, 7)\}$$

$$\therefore h^{-1} = \{(5, 2), (6, 3), (7, 4)\}$$

Question 1(d)

The given function is,

$$n = \{(3, 2), (5, 4), (7, 6)\}$$

$$\therefore n^{-1} = \{(2, 3), (4, 5), (6, 7)\}$$

2. Write down the inverse functions as a set of ordered pairs.

Question 2(a)

From the figure, $d = \{(5, 4), (7, 6), (9, 8)\}$

$\therefore d^{-1} = \{(4, 5), (6, 7), (8, 9)\}$

Question 2(b)

From the figure, $f = \{(1, 3), (2, 4), (3, 5)\}$

$\therefore f^{-1} = \{(3, 1), (4, 2), (5, 3)\}$

3. Find the inverse images from the mapping diagrams.

Question 3(a)

From the mapping diagram,

$g$ = $\{(0, 2), (1, 3), (2, 5)\}$

To find: $g^{-1}\{3, 5\}$

$g(1) = 3 \implies g^{-1}(3) = 1$
$g(2) = 5 \implies g^{-1}(5) = 2$

$\therefore g^{-1}\{3, 5\} = \{1, 2\}$

Question 3(b)

From the mapping diagram,

$i$ is $\{(5, 3), (6, 4), (7, 5)\}$

To find: $i^{-1}\{3\}$

$i(5) = 3 \implies i^{-1}(3) = 5$

$\therefore i^{-1}\{3\} = \{5\}$

4. Find the inverse functions of the following functions.

Question 4(a)

Given:

$f = \{(x, y) : y = x \text{ and } x = \{1, 2, 3\}\}$

$f = \{(1, 1), (2, 2), (3, 3)\}$ [function is defined as identity function, x = y]

By swapping $x$ and $y$

$$\therefore f^{-1} = \{(1, 1), (2, 2), (3, 3)\}$$

Question 4(b)

Given:

$f = \{(x, y) : y = 2x + 1, x = \{0, 1, 2, 3\}\}$

Calculate the pairs for $f$:

$x=0, y=2(0)+1=1 \to (0, 1)$
$x=1, y=2(1)+1=3 \to (1, 3)$
$x=2, y=2(2)+1=5 \to (2, 5)$
$x=3, y=2(3)+1=7 \to (3, 7)$

$$\therefore f = \{(0, 1), (1, 3), (2, 5), (3, 7)\}$$

$$\therefore f^{-1} = \{(1, 0), (3, 1), (5, 2), (7, 3)\}$$

5. Find $f^{-1}$ and the domain and range of $f^{-1}$.

Question 5(a)

Given:

$f = \{(1, 2), (2, 3), (4, 5)\}$

$f^{-1} = \{(2, 1), (3, 2), (5, 4)\}$
Domain of $f^{-1}$ (First elements): $\{2, 3, 5\}$
Range of $f^{-1}$ (Second elements): $\{1, 2, 4\}$

Question 5(b)

Given:

$f = \{(2, 4), (3, 6), (4, 8)\}$

$f^{-1} = \{(4, 2), (6, 3), (8, 4)\}$
Domain of $f^{-1}$: $\{4, 6, 8\}$
Range of $f^{-1}$: $\{2, 3, 4\}$

6. Find the inverse functions and composite mappings.

Question 6(a)

Given:

$f = \{(1, 5), (2, 4), (3, 6)\}$

$g \circ f = \{(5, 6), (4, 5), (6, 4)\}$.

To find $g^{-1}$, we first determine $g$:

From $f$: $f(1)=5, f(2)=4, f(3)=6$.
From $g \circ f$:

$g(f(1))=5 \implies g(5)=6$ [Must be $\textbf{6}$ because input of $g \circ f$ is 5 and out put is 6]

$g(f(2))=5 \implies g(4)=5$

$g(f(3))=4 \implies g(6)=4$

So, $g = \{(5, 6), (4, 5), (6, 4)\}$

$$\therefore g^{-1} = \{(6, 5), (5, 4), (4, 6)\}$$

Question 6(b)

Given:

$f = \{(2, 3), (4, 5), (6, 7)\}$

$g \circ f = \{(2, 4), (4, 6), (6, 8)\}$.

From $f$: $f(2)=3, f(4)=5, f(6)=7$
From $g \circ f$:

$g[f(2)] = g(3) = 4$ [Must be $\textbf{4}$ because input of $g \circ f$ is 2 and output is 4]

$g[f(4)] = g(5) = 6$

$g[f(6)] = g(7) = 8$

So, $g = \{(3, 4), (5, 6), (7, 8)\}$

$$\therefore g^{-1} = \{(4, 3), (6, 5), (8, 7)\}$$

7. Find the inverse functions of the following one-to-one onto functions.

Question 7(a)

$f(x) = 2x + 3$

Let $y = f(x) = 2x + 3$

Interchanging $x$ and $y$

$x = 2y + 3$

or, $x - 3 = 2y$

or, $\frac{x - 3}{2} = y$

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

Question 7(d)

$f(x) = \frac{x - 3}{4}$

Let $y = \frac{x - 3}{4}$

Interchanging $x$ and $y$

$x = \frac{y - 3}{4}$

or, $4x = y - 3$

or, $4x + 3 = y$

$$\therefore f^{-1}(x) = 4x + 3$$

Question 7(g)

$f(x) = \frac{2x + 3}{x - 3}$

Let $y = \frac{2x + 3}{x - 3}$

Interchanging $x$ and $y$

$x = \frac{2y + 3}{y - 3}$

or, $x(y - 3) = 2y + 3$

or, $xy - 3x = 2y + 3$

or, $xy - 2y = 3x + 3$ [both $y$ must be on same side]

or, $y(x - 2) = 3x + 3$

or, $y = \frac{3x + 3}{x - 2}$

$$\therefore f^{-1}(x) = \frac{3x + 3}{x - 2}$$

Question 7(b)

$f(x) = 4x - 5$

Let $y = f(x) = 4x - 5$

Interchanging $x$ and $y$

$x = 4y - 5$

or, $4y = x + 5$

or, $y = \frac{x + 5}{4}$

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

Question 7(c)

$f(x) = \frac{x + 5}{3}$

Let $y = f(x) = \frac{x + 5}{3}$

Interchanging $x$ and $y$

$x = \frac{y + 5}{3}$

or, $3x = y + 5$

or, $3x - 5 = y$

$$\therefore f^{-1}(x) = 3x - 5$$

Question 7(e)

$f: x \to \frac{2x + 5}{6}$

Let $y = f(x) = \frac{2x + 5}{6}$

Interchanging $x$ and $y$

$x = \frac{2y + 5}{6}$

or, $6x = 2y + 5$

or, $2y = 6x - 5$

or, $y = \frac{6x - 5}{2}$

$$\therefore f^{-1}(x) = \frac{6x - 5}{2}$$

Question 7(f)

$g: x \to \frac{3x - 5}{7}$

Let $y = g(x) = \frac{3x - 5}{7}$

Interchanging $x$ and $y$

$x = \frac{3y - 5}{7}$

or, $7x = 3y - 5$

or, $3y = 7x + 5$

or, $y = \frac{7x + 5}{3}$

$$\therefore g^{-1}(x) = \frac{7x + 5}{3}$$

Question 7(h)

$g: x \to \frac{3x - 5}{x + 2}$

Let $y = g(x) = \frac{3x - 5}{x + 2}$

Interchanging $x$ and $y$

$x = \frac{3y - 5}{y + 2}$

or, $x(y + 2) = 3y - 5$

or, $xy + 2x = 3y - 5$

or, $xy - 3y = -2x - 5$

or, $y(x - 3) = -(2x + 5)$

or, $y = \frac{-(2x + 5)}{x - 3}$

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

Question 7(i)

$f: x \to \frac{4x + 7}{5x + 3}$

Let $y = f(x) = \frac{4x + 7}{5x + 3}$

Interchanging $x$ and $y$

$x = \frac{4y + 7}{5y + 3}$

or, $x(5y + 3) = 4y + 7$

or, $5xy + 3x = 4y + 7$

or, $5xy - 4y = 7 - 3x$

or, $y(5x - 4) = 7 - 3x$

or, $y = \frac{7 - 3x}{5x - 4}$

$$\therefore f^{-1}(x) = \frac{7 - 3x}{5x - 4}$$

Question 8(a)

If $f(x) = \frac{x - 5}{4}$ is one - one onto function and $f^{-1}(x) = 25$, find the value of $x$.

Given

$f(x) = \frac{x - 5}{4}$

$f^{-1}(x) = 25$

To find $f^{-1}(x)$:

Let $y = f(x) = \frac{x - 5}{4}$

$x = \frac{y - 5}{4}$

or, $4x = y - 5$

or, $y = 4x + 5$

So, $f^{-1}(x) = 4x + 5$

According to question,

$f^{-1}(x) = 25$

or, $4x + 5 = 25$

or, $4x = 20$.

$$\therefore x = 5$$

Question 8(b)

If $f(x) = 2x - a$ is one - one onto function and $f^{-1}(2) = 2$, find the value of $a$.

Given

$f(x) = 2x - a$

$f^{-1}(2) = 2$

To find: $f^{-1}(x)$

Let $y = f(x) = 2x - a$

Interchanging $x$ and $y$

$x = 2y - a$

or, $x + a = 2y$

or, $\frac{x + a}{2} = y$

$\therefore f^{-1}(x) = \frac{x + a}{2}$

According to question,

$f^{-1}(2) = 2$

or, $\frac{2 + a}{2} = 2$

or, $2 + a = 4$

or, $a = 4 - 2$

$\therefore a = 2$

$\textbf{Question 9(a)}$

$\text{If } f(4x+5)=12x+18 \text{ is one-one onto function, find } f^{-1}(x).$

$\textbf{Solution}$

$\text{Given,}$

$$ f(4x+5)=12x+18 $$

$\text{Let}$

$4x+5 = z$

$4x = z - 5$

$x = \frac{z - 5}{4}$

$\text{Now substitute } x=\frac{z - 5}{4} \text{ in } 12x+18$

$f(z) = 12\left(\frac{z - 5}{4}\right)+18$

or, $f(z)=3(z - 5) + 18$

or, $f(y) = 3y - 15 + 18$

or, $f(z) = 3z + 3$

When $z \rightarrow x$

$$f(x)=3x+3$$

$\text{To find: } f^{-1}(x).$

$\text{Let}$

$f(x)=y = 3x + 3$

Interchanging $x$ and $y$,

$y=3x+3$

or, $y=3x+3$

or, $y-3=3x$

or, $x=\frac{y-3}{3}$

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

$$\boxed{f^{-1}(x)=\frac{x-3}{3}}$$

$\textbf{Question 9(b)}$

$\text{If } f(x+3)=3x+5 \text{ is one-one onto function, find } f^{-1}(x).$

$\textbf{Solution}$

$\text{Given,}$

$$
f(x+3)=3x+5
$$

$\text{Let}$

$$
x+3=y
$$

$\text{Now solve for } x.$

$$
x+3=y
$$

$$
x=y-3
$$

$\text{Now substitute } x=y-3 \text{ in } 3x+5.$

$$
f(y)=3(y-3)+5
$$

$$
f(y)=3y-9+5
$$

$$
f(y)=3y-4
$$

$\text{Therefore,}$

$$
f(x)=3x-4
$$

$\text{Now find } f^{-1}(x).$

$\text{Let}$

$$
f(x)=y
$$

$\text{So,}$

$$
y=3x-4
$$

$\text{Now solve for } x.$

$$
y=3x-4
$$

$$
y+4=3x
$$

$$
x=\frac{y+4}{3}
$$

$\text{Now replace } y \text{ by } x.$

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

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

9. Find $f^{-1}(x)$ from composite inputs.

Question 9(a)

Given $f(4x + 5) = 12x + 18$.

Let $t = 4x + 5 \implies x = \frac{t - 5}{4}$.

$f(t) = 12\left(\frac{t - 5}{4}\right) + 18 = 3(t - 5) + 18 = 3t - 15 + 18 = 3t + 3$.

So, $f(x) = 3x + 3$.

To find $f^{-1}(x)$: Let $y = 3x + 3 \implies x = 3y + 3 \implies y = \frac{x - 3}{3}$.

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

10. Find $f(x+2)$ and $f^{-1}(x+2)$.

Question 10(a)

Given $f(x) = 2x + 7$.

$f(x + 2) = 2(x + 2) + 7 = 2x + 4 + 7 = 2x + 11$
To find $f^{-1}(x)$: Let $y = 2x + 7 \implies x = 2y + 7 \implies y = \frac{x - 7}{2}$.

So, $f^{-1}(x) = \frac{x - 7}{2}$.

$f^{-1}(x + 2) = \frac{(x + 2) - 7}{2} = \frac{x - 5}{2}$

Question 10(b)

Given: $f(x) = 4 - 3x$

Find $f(x+2)$:

$f(x+2) = 4 - 3(x+2) = 4 - 3x - 6$

$f(x+2) = -3x - 2$
Find $f^{-1}(x+2)$:

Let $y = 4 - 3x$. Interchanging $x$ and $y$:

$x = 4 - 3y \implies 3y = 4 - x \implies y = \frac{4-x}{3}$

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

$f^{-1}(x+2) = \frac{4-(x+2)}{3} = \frac{2-x}{3}$

11. Evaluate composite inverse functions.

Question 11(a)

If $f(x) = 2x + 1$ and $g(x) = \frac{x-5}{2}$, find the value of $f^{-1}g^{-1}(3)$.

Find $f^{-1}(x)$: $y = 2x + 1 \implies x = 2y + 1 \implies y = \frac{x-1}{2}$. So, $f^{-1}(x) = \frac{x-1}{2}$.
Find $g^{-1}(x)$: $y = \frac{x-5}{2} \implies x = \frac{y-5}{2} \implies 2x = y - 5 \implies y = 2x + 5$. So, $g^{-1}(x) = 2x + 5$.
Calculate $g^{-1}(3) = 2(3) + 5 = 11$.
Calculate $f^{-1}(11) = \frac{11-1}{2} = \frac{10}{2} = 5$.

Value: $5$

Question 11(b)

If $3.f(x) = 4x + 5$ and $g(x) = 5x - 4$, find the value of $f^{-1} \circ g^{-1}(1)$.

$f(x) = \frac{4x+5}{3}$. Let $y = \frac{4x+5}{3} \implies x = \frac{4y+5}{3} \implies 3x = 4y + 5 \implies y = \frac{3x-5}{4}$. So, $f^{-1}(x) = \frac{3x-5}{4}$.
$g(x) = 5x - 4$. Let $y = 5x - 4 \implies x = 5y - 4 \implies y = \frac{x+4}{5}$. So, $g^{-1}(x) = \frac{x+4}{5}$.
Calculate $g^{-1}(1) = \frac{1+4}{5} = 1$.
Calculate $f^{-1}(1) = \frac{3(1)-5}{4} = -\frac{2}{4} = -\frac{1}{2}$.

Value: $-\frac{1}{2}$

12. Function values and compositions.

Question 12(a)

If $f(x) = 1 + 2x$ and $g(x) = \frac{1}{1-x}$, find $g^{-1}(\frac{1}{2})$ and $f \circ g (-1)$.

Find $g^{-1}(x)$: $y = \frac{1}{1-x} \implies x = \frac{1}{1-y} \implies 1-y = \frac{1}{x} \implies y = 1 - \frac{1}{x}$. So, $g^{-1}(x) = \frac{x-1}{x}$.
$g^{-1}(\frac{1}{2}) = \frac{1/2 - 1}{1/2} = \frac{-1/2}{1/2} = -1$
$g(-1) = \frac{1}{1 - (-1)} = \frac{1}{2}$.
$f \circ g (-1) = f(\frac{1}{2}) = 1 + 2(\frac{1}{2}) = 1 + 1 = 2$

Question 12(b)

$f(x) = \frac{2x + 3}{x + 2}$ and $g(x) = x - 2$.

$f^{-1}(x)$: Let $y = \frac{2x + 3}{x + 2} \implies x = \frac{2y + 3}{y + 2} \implies xy + 2x = 2y + 3 \implies y(x - 2) = 3 - 2x$.

$f^{-1}(x) = \frac{3 - 2x}{x - 2}$
$f^{-1}(1)$: $\frac{3 - 2(1)}{1 - 2} = \frac{1}{-1} = \mathbf{-1}$
$fg(x)$: $f(x - 2) = \frac{2(x - 2) + 3}{(x - 2) + 2} = \frac{2x - 4 + 3}{x} = \mathbf{\frac{2x - 1}{x}}$
$fg(1)$: $\frac{2(1) - 1}{1} = \mathbf{1}$

13. Proofs and specific values.

Question 13(a)

If $f(x) = 3x + 4$ and $g(x) = 2(x + 1)$, prove $f \circ g = g \circ f$ and find $f^{-1}(2)$.

$f \circ g(x) = f(2x+2) = 3(2x+2) + 4 = 6x + 6 + 4 = 6x + 10$.
$g \circ f(x) = g(3x+4) = 2((3x+4) + 1) = 2(3x+5) = 6x + 10$.

Hence, $f \circ g = g \circ f$.
$y = 3x + 4 \implies x = 3y + 4 \implies y = \frac{x-4}{3}$.

$f^{-1}(2) = \frac{2-4}{3} = -\frac{2}{3}$

14. Finding constants and values.

Question 14(a)

Given $f(x) = 3x + a$. If $f(f(6)) = 10$, find $a$ and $f^{-1}(4)$.

$f(6) = 3(6) + a = 18 + a$.
$f(f(6)) = f(18+a) = 3(18+a) + a = 54 + 3a + a = 54 + 4a$.
$54 + 4a = 10 \implies 4a = -44 \implies \mathbf{a = -11}$.
Now $f(x) = 3x - 11$. $y = 3x - 11 \implies x = 3y - 11 \implies y = \frac{x+11}{3}$.

$f^{-1}(4) = \frac{4+11}{3} = \frac{15}{3} = 5$

Question 14(b)

If $f(x) = \frac{x}{2x-3}$ and $f(x) = f^{-1}(x)$, find the value of $x$.

Find $f^{-1}(x)$: $y = \frac{x}{2x-3} \implies x = \frac{y}{2y-3} \implies 2xy - 3x = y \implies 2xy - y = 3x \implies y(2x-1) = 3x$.

So, $f^{-1}(x) = \frac{3x}{2x-1}$.
Set $f(x) = f^{-1}(x)$: $\frac{x}{2x-3} = \frac{3x}{2x-1}$.
$x(2x-1) = 3x(2x-3) \implies 2x^2 - x = 6x^2 - 9x$.
$4x^2 - 8x = 0 \implies 4x(x - 2) = 0$.

$x = 0$ or $x = 2$

Question 15(a)

Given

$f(x) = 4x + 7$

$g(x) = 3x - 5$

$fg^{-1}(x) = 15$

Find $g^{-1}(x) = \frac{x + 5}{3}$.
$f(g^{-1}(x)) = 4(\frac{x + 5}{3}) + 7 = 15$.
$\frac{4x + 20}{3} = 8 \implies 4x + 20 = 24 \implies 4x = 4$.

$x = 1$

Question 15(b)

$f(x) = x^2 - 2x$ and $g(x) = 2x + 3$. If $fg^{-1}(x) = 3$, calculate the value of $x$.

Find $g^{-1}(x)$: Let $y = 2x + 3 \implies x = 2y + 3 \implies y = \frac{x-3}{2}$.
Substitute into $f(x)$: $f\left(\frac{x-3}{2}\right) = \left(\frac{x-3}{2}\right)^2 - 2\left(\frac{x-3}{2}\right) = 3$.
Simplify: $\frac{(x-3)^2}{4} - (x-3) = 3$.
Multiply by 4: $(x-3)^2 - 4(x-3) = 12$.
Let $a = (x-3)$, then $a^2 - 4a - 12 = 0 \implies (a-6)(a+2) = 0$.
$a = 6 \implies x-3 = 6 \implies \mathbf{x = 9}$.
$a = -2 \implies x-3 = -2 \implies \mathbf{x = 1}$.

Question 15(d)

$f(x) = \frac{x}{2-x}$, $g(x) = bx - 2$, and $gf(4) = -8$. Find $f^{-1}(-2)$ and $b$.

Find $f(4)$: $f(4) = \frac{4}{2-4} = \frac{4}{-2} = -2$.
Use $gf(4) = -8$: $g(-2) = b(-2) - 2 = -8 \implies -2b = -6 \implies \mathbf{b = 3}$.
Find $f^{-1}(x)$: Let $y = \frac{x}{2-x} \implies x = \frac{y}{2-y} \implies 2x - xy = y \implies y(1+x) = 2x \implies f^{-1}(x) = \frac{2x}{1+x}$.
$f^{-1}(-2) = \frac{2(-2)}{1+(-2)} = \frac{-4}{-1} = \mathbf{4}$.

Question 15(e)

$f(x) = \frac{3x+11}{x-3}$ and $g(x) = \frac{x-3}{2}$. Find $f^{-1}(x)$ and $x$ if $f(x) = g^{-1}(x)$.

Find $f^{-1}(x)$: Let $y = \frac{3x+11}{x-3} \implies x = \frac{3y+11}{y-3} \implies xy - 3x = 3y + 11 \implies y(x-3) = 3x + 11 \implies \mathbf{f^{-1}(x) = \frac{3x+11}{x-3}}$. (Note: $f$ is its own inverse).
Find $g^{-1}(x)$: $y = \frac{x-3}{2} \implies x = \frac{y-3}{2} \implies 2x = y - 3 \implies \mathbf{g^{-1}(x) = 2x + 3}$.
Solve $f(x) = g^{-1}(x)$: $\frac{3x+11}{x-3} = 2x + 3 \implies 3x + 11 = (2x+3)(x-3)$.
$3x + 11 = 2x^2 - 6x + 3x - 9 \implies 2x^2 - 6x - 20 = 0 \implies x^2 - 3x - 10 = 0$.
$(x-5)(x+2) = 0 \implies \mathbf{x = 5, -2}$.

Question 15(c)

Given $f(x) = 4x - 17$, $g(x) = \frac{2x + 8}{5}$, and $f(x) = g^{-1}(x)$.

Find $g^{-1}(x)$: $x = \frac{2y + 8}{5} \implies 5x = 2y + 8 \implies y = \frac{5x - 8}{2}$.
$4x - 17 = \frac{5x - 8}{2} \implies 8x - 34 = 5x - 8 \implies 3x = 26$.

$x = \frac{26}{3}$

Question 16(a)

$f(x) = \frac{3x + 5}{x - 2}$, $g(x) = 7$. Show $f^{-1} \circ g(x)$ is constant.

$f^{-1}(x) = \frac{2x + 5}{x - 3}$.
$f^{-1}(g(x)) = f^{-1}(7) = \frac{2(7) + 5}{7 - 3} = \frac{19}{4}$.

Since $\frac{19}{4}$ is a number without $x$, it is a constant.

Question 16(b)

$h(x) = \frac{2x-3}{7}$ and $f(x) = 3$. Prove that $h^{-1} \circ f(x)$ is constant.

Find $h^{-1}(x)$: $y = \frac{2x-3}{7} \implies x = \frac{2y-3}{7} \implies 7x = 2y - 3 \implies y = \frac{7x+3}{2}$.
$h^{-1}(f(x)) = h^{-1}(3) = \frac{7(3)+3}{2} = \frac{24}{2} = \mathbf{12}$.
Conclusion: Since $12$ is a constant value independent of $x$, the function is constant.

17. Proof of Quadratic Functions

Question 17(a)

$f(x) = 2x + 3$ and $g(x) = x^2 + 2x + 1$. Prove $f^{-1} \circ g(x)$ is quadratic.

$f^{-1}(x) = \frac{x-3}{2}$.
$f^{-1}(g(x)) = \frac{(x^2 + 2x + 1) - 3}{2} = \mathbf{\frac{x^2 + 2x - 2}{2}}$.
Conclusion: The resulting expression is of degree 2, making it a quadratic function.

Question 18(a)

$f(x) = 3x + 5$, $g(x) = \frac{3x + 2}{4}$. Find $x$ for $f \circ g^{-1}(x)$ to be identity ($x$).

$g^{-1}(x) = \frac{4x - 2}{3}$.
$f(g^{-1}(x)) = 3(\frac{4x - 2}{3}) + 5 = 4x - 2 + 5 = 4x + 3$.
Set $4x + 3 = x \implies 3x = -3$.

$x = -1$

Question 18(b)

$f(x) = 4x + 2$ and $g(x) = \frac{4x+3}{4}$. Find $x$ such that $f \circ g^{-1}(x)$ is an identity function.

Find $g^{-1}(x)$: $y = \frac{4x+3}{4} \implies x = \frac{4y+3}{4} \implies 4x = 4y + 3 \implies y = \frac{4x-3}{4}$.
$f(g^{-1}(x)) = 4\left(\frac{4x-3}{4}\right) + 2 = 4x - 3 + 2 = 4x - 1$.
Set $f(g^{-1}(x)) = x$ (Identity condition): $4x - 1 = x \implies 3x = 1$.
$\mathbf{x = \frac{1}{3}}$.

Question 19

(a) $f = \{(1, 3), (2, 5), (3, 7)\} \implies \mathbf{f^{-1} = \{(3, 1), (5, 2), (7, 3)\}}$

(b) $\mathbf{f^{-1}(5) = 2}$

(d) Let $(g \circ f)^{-1}(x) = 40$. (Assuming a linear pattern $y = 6x + 2$ for $g \circ f$):

$(g \circ f)^{-1}(x) = \frac{x - 2}{6} = 40 \implies x - 2 = 240 \implies \mathbf{x = 242}$.