Inverse Functions and Graphs of Simple Functions

Understanding Inverse Functions

Before we dive into inverse functions, let's recall what a function is. A function maps elements from one set (the domain) to another set (the codomain). For a function to have an inverse, it must be a bijective function, meaning it must be both one-to-one (injective) and onto (surjective).

A function is one-to-one if each element in the codomain is mapped to by at most one element in the domain. In simpler terms, different inputs always produce different outputs. Graphically, this means any horizontal line drawn on the graph of the function will intersect the graph at most once (the Horizontal Line Test).

A function is onto if every element in the codomain is mapped to by at least one element in the domain. This means the range of the function is equal to its codomain.

If a function $f$ is bijective, its inverse function, denoted as $f^{-1}$, reverses the mapping. If $f(a) = b$, then $f^{-1}(b) = a$. The domain of $f^{-1}$ is the range of $f$, and the range of $f^{-1}$ is the domain of $f$.

To find the inverse of a function $y = f(x)$, we follow these steps:

1. Replace $f(x)$ with $y$.
2. Swap $x$ and $y$.
3. Solve the new equation for $y$.
4. Replace $y$ with $f^{-1}(x)$.

For example, let's find the inverse of $f(x) = 2x + 3$.

Step 1: $y = 2x + 3$

Step 2: $x = 2y + 3$

Step 3: $x - 3 = 2y$ $y = \frac{x - 3}{2}$

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

Graphs of Simple Functions and Their Inverses

Understanding the graphs of basic functions is crucial for visualizing their inverses and for analyzing their properties. Let's look at some common simple functions and their inverses.

1. Linear Functions

A linear function is of the form $f(x) = mx + c$, where $m \neq 0$. These functions are always bijective (one-to-one and onto) and thus always have an inverse.

Let's find the inverse of $f(x) = mx + c$:

$y = mx + c$ $x = my + c$ $x - c = my$ $y = \frac{x - c}{m}$

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

Example: $f(x) = 3x - 1$. $y = 3x - 1$ $x = 3y - 1$ $x + 1 = 3y$ $y = \frac{x + 1}{3}$ $f^{-1}(x) = \frac{x + 1}{3}$.

The graph of $y = mx + c$ is a straight line with slope $m$ and y-intercept $c$. The graph of its inverse $y = \frac{1}{m}x - \frac{c}{m}$ is also a straight line. Both graphs are reflections of each other across the line $y = x$.

2. Quadratic Functions (Restricted Domain)

A standard quadratic function $f(x) = x^2$ is not one-to-one because, for example, $f(2) = 4$ and $f(-2) = 4$. To define an inverse, we must restrict its domain to make it one-to-one.

Consider $f(x) = x^2$ with the domain $x \ge 0$. This function is one-to-one and onto (with range $y \ge 0$).

To find its inverse:

$y = x^2$ $x = y^2$ $y = \pm \sqrt{x}$

Since the original domain was $x \ge 0$, the range of the inverse must be $y \ge 0$. Therefore, we choose the positive square root: $f^{-1}(x) = \sqrt{x}$. The domain of $f^{-1}(x)$ is $x \ge 0$.

If we considered $f(x) = x^2$ with the domain $x \le 0$, then the inverse would be $f^{-1}(x) = -\sqrt{x}$ (with domain $x \ge 0$).

The graph of $y = x^2$ for $x \ge 0$ is the right half of a parabola opening upwards. Its inverse, $y = \sqrt{x}$, is the upper half of a parabola opening to the right. Both are reflections across $y = x$.

3. Cubic Functions

Consider the function $f(x) = x^3$. This function is one-to-one and onto for all real numbers.

To find its inverse:

$y = x^3$ $x = y^3$ $y = \sqrt[3]{x}$

So, $f^{-1}(x) = \sqrt[3]{x}$. The domain and range of both $f(x)$ and $f^{-1}(x)$ are all real numbers.

The graph of $y = x^3$ has an 'S' shape, passing through the origin. The graph of $y = \sqrt[3]{x}$ is also 'S' shaped but oriented differently, and it too passes through the origin. They are reflections of each other across the line $y = x$.

4. Exponential Functions

Consider the function $f(x) = a^x$, where $a > 0$ and $a \neq 1$. This function is one-to-one and onto (with range $y > 0$).

To find its inverse:

$y = a^x$ $x = a^y$

To solve for $y$, we use the definition of logarithms: $y = \log_a x$.

So, $f^{-1}(x) = \log_a x$. The domain of $f^{-1}(x)$ is $x > 0$, and its range is all real numbers.

Example: $f(x) = 2^x$. Its inverse is $f^{-1}(x) = \log_2 x$.

The graph of $y = a^x$ is an increasing (if $a > 1$) or decreasing (if $0 < a < 1$) curve that passes through $(0, 1)$ and approaches the x-axis asymptotically. The graph of $y = \log_a x$ is a curve that passes through $(1, 0)$ and approaches the y-axis asymptotically. They are reflections across $y = x$.

5. Logarithmic Functions

The logarithmic function $f(x) = \log_a x$ (where $a > 0$ and $a \neq 1$) is the inverse of the exponential function $a^x$. It is one-to-one and onto (with domain $x > 0$).

To find its inverse:

$y = \log_a x$ $x = \log_a y$

Converting to exponential form: $y = a^x$.

So, $f^{-1}(x) = a^x$. The domain of $f^{-1}(x)$ is all real numbers, and its range is $y > 0$.

Example: $f(x) = \log_{10} x$ (common logarithm). Its inverse is $f^{-1}(x) = 10^x$.

6. Trigonometric Functions (Restricted Domain)

Standard trigonometric functions like $\sin x$, $\cos x$, and $\tan x$ are periodic and therefore not one-to-one over their entire domains. To define their inverses, we restrict their domains to principal value intervals.

• Sine Function: $f(x) = \sin x$. Restricted domain: $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Range: $[-1, 1]$. Inverse: $f^{-1}(x) = \arcsin x$ (or $\sin^{-1} x$). Domain: $[-1, 1]$. Range: $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
• Cosine Function: $f(x) = \cos x$. Restricted domain: $[0, \pi]$. Range: $[-1, 1]$. Inverse: $f^{-1}(x) = \arccos x$ (or $\cos^{-1} x$). Domain: $[-1, 1]$. Range: $[0, \pi]$.
• Tangent Function: $f(x) = \tan x$. Restricted domain: $(-\frac{\pi}{2}, \frac{\pi}{2})$. Range: $(-\infty, \infty)$. Inverse: $f^{-1}(x) = \arctan x$ (or $\tan^{-1} x$). Domain: $(-\infty, \infty)$. Range: $(-\frac{\pi}{2}, \frac{\pi}{2})$.

The graphs of these inverse trigonometric functions are reflections of the restricted portions of the original trigonometric functions across the line $y = x$.

Graphical Analysis of Inverse Functions

The graph of an inverse function $f^{-1}(x)$ is obtained by reflecting the graph of $f(x)$ across the line $y = x$. This is because if $(a, b)$ is a point on the graph of $f(x)$, then $f(a) = b$, which implies $f^{-1}(b) = a$. Thus, $(b, a)$ is a point on the graph of $f^{-1}(x)$. Swapping the coordinates $(a, b)$ to $(b, a)$ geometrically corresponds to a reflection across the line $y = x$.

Key Points for Graphical Analysis:

• If $f(x)$ is increasing, $f^{-1}(x)$ is also increasing.
• If $f(x)$ is decreasing, $f^{-1}(x)$ is also decreasing.
• The domain of $f(x)$ becomes the range of $f^{-1}(x)$, and the range of $f(x)$ becomes the domain of $f^{-1}(x)$.
• Points where $f(x)$ intersects $y=x$ are fixed points for the inverse transformation.

Example: Graphing $f(x) = \sqrt{x-2}$ and its inverse.

First, find the inverse function. Let $y = \sqrt{x-2}$. The domain is $x-2 \ge 0 \implies x \ge 2$. The range is $y \ge 0$.

Swap $x$ and $y$: $x = \sqrt{y-2}$. Square both sides: $x^2 = y-2$. Solve for $y$: $y = x^2 + 2$.

The domain of the original function ($x \ge 2$) becomes the range of the inverse, so $y \ge 2$. The range of the original function ($y \ge 0$) becomes the domain of the inverse, so $x \ge 0$. Thus, the inverse function is $f^{-1}(x) = x^2 + 2$, with the domain $x \ge 0$.

Graphing:

• The graph of $f(x) = \sqrt{x-2}$ starts at $(2, 0)$ and curves upwards to the right.
• The graph of $f^{-1}(x) = x^2 + 2$ for $x \ge 0$ starts at $(0, 2)$ and curves upwards to the right, forming the right half of a parabola.
• Both graphs are reflections of each other across the line $y = x$. For instance, the point $(6, 2)$ is on $f(x)$ since $\sqrt{6-2} = \sqrt{4} = 2$. The point $(2, 6)$ is on $f^{-1}(x)$ since $2^2 + 2 = 4 + 2 = 6$.

Properties of Inverse Functions

For a bijective function $f$ with inverse $f^{-1}$:

• $f(f^{-1}(x)) = x$ for all $x$ in the domain of $f^{-1}$.
• $f^{-1}(f(x)) = x$ for all $x$ in the domain of $f$.

These are known as the inverse function identities. They are fundamental to verifying if two functions are indeed inverses of each other.

Example: Verify that $f(x) = 3x - 1$ and $g(x) = \frac{x+1}{3}$ are inverses.

Check $f(g(x))$: $f\left(\frac{x+1}{3}\right) = 3\left(\frac{x+1}{3}\right) - 1 = (x+1) - 1 = x$.

Check $g(f(x))$: $g(3x - 1) = \frac{(3x - 1) + 1}{3} = \frac{3x}{3} = x$.

Since both compositions result in $x$, $f(x)$ and $g(x)$ are indeed inverse functions.

Inverse of Composite Functions

If $f$ and $g$ are invertible functions, then their composition $(f \circ g)(x) = f(g(x))$ is also invertible. The inverse of a composition follows a specific rule:

$(f \circ g)^{-1}(x) = (g^{-1} \circ f^{-1})(x)$

This means the inverse of a composition of two functions is the composition of their inverses in the reverse order.

Proof Sketch: Let $h(x) = (f \circ g)(x)$. We want to find $h^{-1}(x)$. Let $y = h(x) = f(g(x))$. Apply $f^{-1}$ to both sides: $f^{-1}(y) = f^{-1}(f(g(x)))$. Using the inverse identity, $f^{-1}(y) = g(x)$. Now apply $g^{-1}$ to both sides: $g^{-1}(f^{-1}(y)) = g^{-1}(g(x))$. Using the inverse identity again, $g^{-1}(f^{-1}(y)) = x$. So, $x = g^{-1}(f^{-1}(y))$. Since $y = h(x)$, we have $x = h^{-1}(y)$. Therefore, $h^{-1}(y) = g^{-1}(f^{-1}(y))$. Replacing $y$ with $x$, we get $h^{-1}(x) = (g^{-1} \circ f^{-1})(x)$.

Generalization: For a composition of three functions, $(f \circ g \circ h)^{-1} = h^{-1} \circ g^{-1} \circ f^{-1}$. The order is reversed.

Example: Let $f(x) = x+1$ and $g(x) = 2x$. Find $(f \circ g)^{-1}(x)$. First, find $f^{-1}(x)$ and $g^{-1}(x)$. $f(x) = x+1 \implies f^{-1}(x) = x-1$. $g(x) = 2x \implies g^{-1}(x) = \frac{x}{2}$.

Using the rule: $(f \circ g)^{-1}(x) = (g^{-1} \circ f^{-1})(x)$. $(g^{-1} \circ f^{-1})(x) = g^{-1}(f^{-1}(x)) = g^{-1}(x-1) = \frac{x-1}{2}$.

Alternatively, first find $(f \circ g)(x)$: $(f \circ g)(x) = f(g(x)) = f(2x) = 2x + 1$. Now find the inverse of $h(x) = 2x + 1$. $y = 2x + 1$ $x = 2y + 1$ $x - 1 = 2y$ $y = \frac{x-1}{2}$. So, $(f \circ g)^{-1}(x) = \frac{x-1}{2}$. The results match.

Summary Table of Simple Functions and Their Inverses

Function $f(x)$	Domain of $f$	Range of $f$	Inverse $f^{-1}(x)$	Domain of $f^{-1}$	Range of $f^{-1}$
$mx + c$ ($m \neq 0$)	$\mathbb{R}$	$\mathbb{R}$	$\frac{x-c}{m}$	$\mathbb{R}$	$\mathbb{R}$
$x^2$ ($x \ge 0$)	$[0, \infty)$	$[0, \infty)$	$\sqrt{x}$	$[0, \infty)$	$[0, \infty)$
$x^3$	$\mathbb{R}$	$\mathbb{R}$	$\sqrt[3]{x}$	$\mathbb{R}$	$\mathbb{R}$
$a^x$ ($a>0, a \neq 1$)	$\mathbb{R}$	$(0, \infty)$	$\log_a x$	$(0, \infty)$	$\mathbb{R}$
$\log_a x$ ($a>0, a \neq 1$)	$(0, \infty)$	$\mathbb{R}$	$a^x$	$\mathbb{R}$	$(0, \infty)$
$\sin x$ ($[-\frac{\pi}{2}, \frac{\pi}{2}]$)	$[-\frac{\pi}{2}, \frac{\pi}{2}]$	$[-1, 1]$	$\arcsin x$	$[-1, 1]$	$[-\frac{\pi}{2}, \frac{\pi}{2}]$
$\cos x$ ($[0, \pi]$)	$[0, \pi]$	$[-1, 1]$	$\arccos x$	$[-1, 1]$	$[0, \pi]$
$\tan x$ ($(-\frac{\pi}{2}, \frac{\pi}{2})$)	$(-\frac{\pi}{2}, \frac{\pi}{2})$	$(-\infty, \infty)$	$\arctan x$	$(-\infty, \infty)$	$(-\frac{\pi}{2}, \frac{\pi}{2})$