Question Statement

Determine whether the given function is one-to-one by examining its graph. If the function is one-to-one, find its inverse. Also, draw the graphs of the inverse function:

(i) $f (x) = \frac{1}{3} x + 3$ (ii) $g (x) = x (x - 5)$ (iii) $h (x) = (x + 1)^{2}$ (iv) $f (x) = x^{3} - 8$ (v) $g (x) = \frac{4}{x}$ (vi) $h (x) = \frac{1}{3 x + 5}$ (vii) $f (x) = x^{4} + 2$ (viii) $g (x) = 5$ (ix) $h (x) = ∣ x ∣$

Background and Explanation

A function is one-to-one (1-1) if every output value corresponds to exactly one input value. Graphically, this is determined by the Horizontal Line Test: if any horizontal line intersects the graph more than once, the function is not one-to-one. Only one-to-one functions have inverse functions, which are found by swapping the roles of $x$ and $y$ and solving for the new $y$ .

Solution

Part (i) $f (x) = \frac{1}{3} x + 3$

First, we construct a table of values to plot the function:

$x$	-6	-3	0	3	6	$\dots$
$f (x)$	1	2	3	4	5	$\dots$

From the graph, it is clear that the function is $1 - 1$ because it passes the horizontal line test. To find its inverse, let $y = f (x)$ :

$y y - 3 \frac{1}{3} x x = \frac{1}{3} x + 3 = \frac{1}{3} x = y - 3 = 3 (y - 3)$

Replace $y$ and $x$ to get the inverse function: $y f^{- 1} (x) = 3 (x - 3) = 3 (x - 3)$

Table for $f^{- 1} (x)$ :

$x$	$\dots$	1	3	4	$\dots$
$f^{- 1} (x)$	$\dots$	-6	0	3	$\dots$

Part (ii) $g (x) = x (x - 5)$

We examine the values of the function:

$x$	$\dots$	-3	-2	-1	0	1	2	3	4	5	$\dots$
$g (x)$	$\dots$	24	14	6	0	-4	-6	-6	-4	0	$\dots$

From the graph and the table, it is clear that it is not a 1-1 function. For example, when $x = 0$ or $x = 5$ , the value of $g (x) = 0$ . Since multiple $x$ values produce the same $y$ value, it fails the horizontal line test. Inverse is not possible.

Part (iii) $h (x) = (x + 1)^{2}$

We examine the values of the function:

$x$	$\dots$	-3	-2	-1	0	1	2	3	$\dots$
$h (x)$	$\dots$	4	1	0	1	4	9	16	$\dots$

From the graph, it is clear that the function is not 1-1 (e.g., $h (- 3) = 4$ and $h (1) = 4$ ). Inverse is not possible.

Part (iv) $f (x) = x^{3} - 8$

$x$	$\dots$	-2	-1	0	1	2	3	$\dots$
$f (x)$	$\dots$	-16	-9		-7	0	19	$\dots$

It is clear from the graph that $f$ is a 1-1 function.

To find $f^{- 1} (x)$ : Let $f (x) = y \Rightarrow x = f^{- 1} (y)$ $y x^{3} x f^{- 1} (y) = x^{3} - 8 = y + 8 = (y + 8)^{1/3} = (y + 8)^{1/3}$

Replace $y$ by $x$ : $f^{- 1} (x) f^{- 1} (x) = (y + x)^{1/3} = (x + 8)^{1/3}$

Table for $f^{- 1} (x)$ :

$x$	$\dots$	-16	-8	0	19	56	$\dots$
$f^{- 1} (x)$	$\dots$	-2	0	2	3	4	$\dots$

Part (v) $g (x) = \frac{4}{x}$

$x$	$\dots$	-8	-4	-2	-1	0	1	2	4	8	$\dots$
$g (x)$	$\dots$	-0.5	-1	-2	-4	$\infty$	4	2	1	0.5	$\dots$

From the graph, it is clear that the function is 1-1. To find the inverse, put $g (x) = y$ : $y x y x = \frac{4}{x} = 4 = \frac{4}{y}$

Replace $x$ with $y$ : $y g^{- 1} (x) = \frac{4}{x} = \frac{4}{x}$ The graph of the inverse is the same as $g (x)$ .

Part (vi) $h (x) = \frac{1}{3 x + 5}$

$x$	$\dots$	-3	-2	-1	0	1	3	5	$\dots$
$h (x)$	$\dots$	-0.25	-1	0.5	0.2	0.125	$\dots$

From the graph, it is observed that the function is 1-1. To find its inverse, let $y = h (x)$ : $y (3 x + 5) y 3 x + 5 3 x x = \frac{1}{3 x + 5} = 1 = \frac{1}{y} = \frac{1}{y} - 5 = \frac{1 - 5 y}{y} = \frac{1 - 5 y}{3 y}$

Replace $y$ with $x$ : $y = \frac{1 - 5 x}{3 x} \Rightarrow h^{- 1} (x) = \frac{1 - 5 x}{3 x}$

Table for $h^{- 1} (x)$ :

$x$	$\dots$	-4	-1	0	2	5	$\dots$
$h^{- 1} (x)$	$\dots$	-1.75	-2	$\infty$	-1.5	-1.6	$\dots$

Part (vii) $f (x) = x^{4} + 2$

$x$	$\dots$	-4	-3	-2	-1	0	1	2	3	4	$\dots$
$f (x)$	$\dots$	258	83	18	3	2	3	18	83	258	$\dots$

From the graph, it is clear that the function is not 1-1 (it is symmetric about the y-axis). Inverse is not possible.

Part (viii) $g (x) = 5$

$x$	$\dots$	-3	-2	-1	0	1	2	3	$\dots$
$g (x)$	$\dots$	5	5	5	5	5	5	5	$\dots$

This is a constant function (a horizontal line). It fails the horizontal line test because a horizontal line at $y = 5$ intersects the graph at infinitely many points. Inverse is not possible.

Part (ix) $h (x) = ∣ x ∣$

$x$	$\dots$	-3	-2	-1	0	1	2	3	$\dots$
$h (x)$	$\dots$	3	2	1	0	1	2	3	$\dots$

From the graph, it is clear that the function is not 1-1 (e.g., $h (- 1) = 1$ and $h (1) = 1$ ). Inverse does not exist.

Key Formulas or Methods Used

Horizontal Line Test: A function is one-to-one if no horizontal line intersects its graph more than once.
Inverse Function Algebra:
1. Set $y = f (x)$ .
2. Solve the equation for $x$ in terms of $y$ .
3. Interchange $x$ and $y$ to find $f^{- 1} (x)$ .
Symmetry: The graph of $f^{- 1} (x)$ is the reflection of the graph of $f (x)$ across the line $y = x$ .

Summary of Steps

Create a Table of Values: Calculate $y$ for various $x$ values to understand the function's behavior.
Graph the Function: Plot the points to visualize the curve.
Check 1-1 Status: Apply the horizontal line test. If it fails, state that the inverse does not exist.
Find the Inverse Algebraically: If 1-1, solve for $x$ and swap variables.
Plot the Inverse: Create a table for the inverse function (by swapping $x$ and $y$ coordinates from the original table) and draw the graph.

Question Statement

Determine whether the given function is one-to-one by examining its graph. If the function is one-to-one, find its inverse. Also, draw the graphs of the inverse function:

Background and Explanation

Solution

Part (i) $f (x) = \frac{1}{3} x + 3$

First, we construct a table of values to plot the function:

$x$	-6	-3	0	3	6	$\dots$
$f (x)$	1	2	3	4	5	$\dots$

From the graph, it is clear that the function is $1 - 1$ because it passes the horizontal line test. To find its inverse, let $y = f (x)$ :

$y y - 3 \frac{1}{3} x x = \frac{1}{3} x + 3 = \frac{1}{3} x = y - 3 = 3 (y - 3)$

Replace $y$ and $x$ to get the inverse function: $y f^{- 1} (x) = 3 (x - 3) = 3 (x - 3)$

Table for $f^{- 1} (x)$ :

$x$	$\dots$	1	3	4	$\dots$
$f^{- 1} (x)$	$\dots$	-6	0	3	$\dots$

Part (ii) $g (x) = x (x - 5)$

We examine the values of the function:

$x$	$\dots$	-3	-2	-1	0	1	2	3	4	5	$\dots$
$g (x)$	$\dots$	24	14	6	0	-4	-6	-6	-4	0	$\dots$

Part (iii) $h (x) = (x + 1)^{2}$

We examine the values of the function:

$x$	$\dots$	-3	-2	-1	0	1	2	3	$\dots$
$h (x)$	$\dots$	4	1	0	1	4	9	16	$\dots$

From the graph, it is clear that the function is not 1-1 (e.g., $h (- 3) = 4$ and $h (1) = 4$ ). Inverse is not possible.

Part (iv) $f (x) = x^{3} - 8$

$x$	$\dots$	-2	-1	0	1	2	3	$\dots$
$f (x)$	$\dots$	-16	-9		-7	0	19	$\dots$

It is clear from the graph that $f$ is a 1-1 function.

To find $f^{- 1} (x)$ : Let $f (x) = y \Rightarrow x = f^{- 1} (y)$ $y x^{3} x f^{- 1} (y) = x^{3} - 8 = y + 8 = (y + 8)^{1/3} = (y + 8)^{1/3}$

Replace $y$ by $x$ : $f^{- 1} (x) f^{- 1} (x) = (y + x)^{1/3} = (x + 8)^{1/3}$

Table for $f^{- 1} (x)$ :

$x$	$\dots$	-16	-8	0	19	56	$\dots$
$f^{- 1} (x)$	$\dots$	-2	0	2	3	4	$\dots$

Part (v) $g (x) = \frac{4}{x}$

$x$	$\dots$	-8	-4	-2	-1	0	1	2	4	8	$\dots$
$g (x)$	$\dots$	-0.5	-1	-2	-4	$\infty$	4	2	1	0.5	$\dots$

From the graph, it is clear that the function is 1-1. To find the inverse, put $g (x) = y$ : $y x y x = \frac{4}{x} = 4 = \frac{4}{y}$

Replace $x$ with $y$ : $y g^{- 1} (x) = \frac{4}{x} = \frac{4}{x}$ The graph of the inverse is the same as $g (x)$ .

Part (vi) $h (x) = \frac{1}{3 x + 5}$

$x$	$\dots$	-3	-2	-1	0	1	3	5	$\dots$
$h (x)$	$\dots$	-0.25	-1	0.5	0.2	0.125	$\dots$

Replace $y$ with $x$ : $y = \frac{1 - 5 x}{3 x} \Rightarrow h^{- 1} (x) = \frac{1 - 5 x}{3 x}$

Table for $h^{- 1} (x)$ :

$x$	$\dots$	-4	-1	0	2	5	$\dots$
$h^{- 1} (x)$	$\dots$	-1.75	-2	$\infty$	-1.5	-1.6	$\dots$

Part (vii) $f (x) = x^{4} + 2$

$x$	$\dots$	-4	-3	-2	-1	0	1	2	3	4	$\dots$
$f (x)$	$\dots$	258	83	18	3	2	3	18	83	258	$\dots$

From the graph, it is clear that the function is not 1-1 (it is symmetric about the y-axis). Inverse is not possible.

Part (viii) $g (x) = 5$

$x$	$\dots$	-3	-2	-1	0	1	2	3	$\dots$
$g (x)$	$\dots$	5	5	5	5	5	5	5	$\dots$

This is a constant function (a horizontal line). It fails the horizontal line test because a horizontal line at $y = 5$ intersects the graph at infinitely many points. Inverse is not possible.

Part (ix) $h (x) = ∣ x ∣$

$x$	$\dots$	-3	-2	-1	0	1	2	3	$\dots$
$h (x)$	$\dots$	3	2	1	0	1	2	3	$\dots$

From the graph, it is clear that the function is not 1-1 (e.g., $h (- 1) = 1$ and $h (1) = 1$ ). Inverse does not exist.

Key Formulas or Methods Used

Horizontal Line Test: A function is one-to-one if no horizontal line intersects its graph more than once.
Inverse Function Algebra:
1. Set $y = f (x)$ .
2. Solve the equation for $x$ in terms of $y$ .
3. Interchange $x$ and $y$ to find $f^{- 1} (x)$ .
Symmetry: The graph of $f^{- 1} (x)$ is the reflection of the graph of $f (x)$ across the line $y = x$ .

Summary of Steps

Create a Table of Values: Calculate $y$ for various $x$ values to understand the function's behavior.
Graph the Function: Plot the points to visualize the curve.
Check 1-1 Status: Apply the horizontal line test. If it fails, state that the inverse does not exist.
Find the Inverse Algebraically: If 1-1, solve for $x$ and swap variables.
Plot the Inverse: Create a table for the inverse function (by swapping $x$ and $y$ coordinates from the original table) and draw the graph.

1.4 Q-2

Question Statement

Background and Explanation

Solution

Part (i) $f (x) = \frac{1}{3} x + 3$

Part (ii) $g (x) = x (x - 5)$

Part (iii) $h (x) = (x + 1)^{2}$

Part (iv) $f (x) = x^{3} - 8$

Part (v) $g (x) = \frac{4}{x}$

Part (vi) $h (x) = \frac{1}{3 x + 5}$

Part (vii) $f (x) = x^{4} + 2$

Part (viii) $g (x) = 5$

Part (ix) $h (x) = ∣ x ∣$

Key Formulas or Methods Used

Summary of Steps

1.4 Q-2

Question Statement

Background and Explanation

Solution

Part (i) $f (x) = \frac{1}{3} x + 3$

Part (ii) $g (x) = x (x - 5)$

Part (iii) $h (x) = (x + 1)^{2}$

Part (iv) $f (x) = x^{3} - 8$

Part (v) $g (x) = \frac{4}{x}$

Part (vi) $h (x) = \frac{1}{3 x + 5}$

Part (vii) $f (x) = x^{4} + 2$

Part (viii) $g (x) = 5$

Part (ix) $h (x) = ∣ x ∣$

Key Formulas or Methods Used

Summary of Steps