Inverse Functions

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one-to-one

Just like numbers, functions have an arithmetic. They have all of the usual number operations, plus one more. Functions also have composition as an operation.

Just like number operations have identity numbers, $0$ and $1$, composition has an identity function - the identity function: $Id(x) = x$.

Around these identity elements, we define inverses.

The additive inverse of $a$ is the number that when added to $a$ results in $0$. Our symbol is $-a$, which we call the opposite of $a$.
The multiplicative inverse of $a$ is the number that when multiplied to $a$ results in $1$. Our symbol is $\frac {1}{a}$ or $a^{-1}$, which we call the reciprocal of $a$.
The composition inverse of a function $f$ is the function that when composed with $f$ gives the identity function. Our symbol is $f^{-1}$, which we call the inverse of $f$.

The Inverse of a Function

Let $f$ be a one-to-one function with domain $D$, range $R$, and pairs $(d, f(d))$.

Then the inverse function of $f$ is a function with domain $R$, range $D$, and pairs $(f(d), d)$.

The inverse of $f$ is denote as $f^{-1}$.

Notation Warning: We reuse and overuse notation in Mathematics. This is an example. Sometimes $f^{-1}$ means reciprocal and sometimes $f^{-1}$ means inverse function. The correct interpretation depends on the context in which it used.

Inverse Functions

Composition glues function pairs together producing a third function.

Let $In$ and $Out$ be two functions and $Out \circ In$ their composition.

Pairs in $In$ look like $(a, In(a))$. The composition then interprets $In(a)$ as a member of the domain of $Out$. $Out$ connects $In(a)$ to $Out(In(a))$. The composition, $Out \circ In$ just ignores the middle part. The composition sees $a$ as a domain number and then $Out(In(a))$ as the range partner.

Suppose we are in a situation where $Out(In(a)) = a$, which means the composition is the identity function. That would tell us that $Out$ is the inverse function of $In$. $Out = In^{-1}$.

Then we would have pairs inside $In$ looking like $(a, In(a))$ and pairs inside $In^{-1}$ looking like $(In(a), a)$. The inverse function just reverses the order pair.

$\blacktriangleright $ The inverse function is just the reverse of the original function.

This brings up an important issue.

Inverse Functions

Reversing the pairs of a function does not always produce a function.

Squaring Function

The squaring function has the formula $Sq(a) =a^2$ with $\mathbb {R}$ as its domain.

It contains the two pairs: $(2, 4)$ and $(-2, 4)$.

If we reverse these pairs, then the inverse function would contain the pairs: $(4, 2)$ and $(4,-2)$.

This would violate the function rule. We would have a domain number in two pairs.

Just like with numbers, some functions don’t have inverses. Like, $0$ just doesn’t have a multiplicative inverse.

One-to-One

Some functions just don’t have an inverse, because they have a range number in two pairs. This is fine for the function, but when reversing the pairs to obtain the inverse funciton, then that range number turns into a domain number in two pairs.

For a function to have an inverse, every range number must be in only one pair.

One-to-One

A function is said to be one-to-one if each range number is in exactly one pair.

Or, an equivalent definition:

One-to-One

A function, $f$, is said to be one-to-one if

\[ f(a) = f(b) \rightarrow a = b \]

This definition says that if two function values are equal, then they couldn’t have come from two different domain numbers. They came from the same domain number. So, each function value is in exactly one pair.

Linear Functions

If $C$ is a constant function, then it is not one-to-one and thus doesn’t have an inverse.

If $C$ is a nonconstant linear function, then it has a formula like $C(x) = a \cdot x + b$ where $a \ne 0$.

Then $C^{-1}(t) = \frac {1}{a} \cdot t - \frac {b}{a}$

The composition is

\[ (C \circ C^{-1})(w) = C(C^{-1})(w) = a \cdot \left (\frac {1}{a} \cdot w - \frac {b}{a}\right ) + b = w \]

and the other way

\[ (C^{-1} \circ C)(y) = C^{-1}(C(y)) = \frac {1}{a} \cdot (a \cdot y + b) - \frac {b}{a} = y \]

Square

The squaring function has the formula $Sq(k) = k^2$ with $\mathbb {R}$ as its domain. It doesn’t have an inverse.

The graph contains the two points: $(2, 4)$ and $(-2, 4)$.

They are at the same vertical height, which means they represent the same function value.

The two points lie on the same horizontal line.

Horizontal Line Test

A function’s graph passes the horizontal line test if there are no horizontal lines that intersect the graph more than once.

A function is one-to-one if its graph passes the horizontal line test.

Salvaging

Functions that are not one-to-one do not have an inverse.

We can’t fix this situation, but we can salvage pieces.

Squaring Function

The squaring function has the formula $Sq(k) = k^2$ with $\mathbb {R}$ as its domain. It doesn’t have an inverse.

However, we can get a partial inverse, by restricting the domain.

The implied domain of $Sq(k) = k^2$ is $\mathbb {R}$. We will restrict this to just the nonnegative numbers: $[0, \infty )$.

Now, it has the square root function as an inverse.

Since the inverse function has the reverse pairs of the original function, its graph is a reflection across the graph of the identity function: $y = Id(k) = k$.

Inverse Function

Let $H(t) = 3t - 5$.

Then $H^{-1}(x)$ is a function such that $H \circ H^{-1} = Id$

We are looking for a function, $H^{-1}$, such that $(H \circ H^{-1})(k) = H(H^{-1}(k)) = 3 H^{-1}(k) - 5 = k$

$H$ is linear and $Id$ is linear, therefore $H^{-1}$ should be linear.

Let’s start building. $H^{-1}(x)$ should look like $a \cdot x + b$. The leading coefficient of $H^{-1}$, $a$, should be $\frac {1}{3}$.

$H^{-1}(x) =\frac {1}{3} x + b$

That would gets us $H(H^{-1}(k)) = 3 (\frac {1}{3}x + b) - 5 = x + 3b - 5 $

We need $3b - 5 = 0$ or $b = \answer {\frac {5}{3}}$.

$H^{-1}(x) =\frac {1}{3} x + \frac {5}{3}$

Inverse

Suppose $T$ is a one-to-one function with $T(4) = 5$. Then which inverse value do we know?

$T^{-1}(-4) = -5$ $T^{-1}\left (\frac {1}{4}\right ) = \frac {1}{5}$ $T^{-1}(-5) = -4$ $T^{-1}(5) = 4$

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2025-01-07 01:06:20