Inverse of a Function
Inverse of a Function
Consider a function \( f : A \to B \), and we wish to define a new function \( g : B \to A \) that "undoes" the operation of \( f \). Specifically, if \( f(a) = b \), then \( g(b) = a \). Such a function \( g \) is called the inverse of \( f \). Intuitively, \( g \) reverses the mapping of \( f \), taking the outputs of \( f \) and assigning them back to their corresponding inputs. However, constructing such a function \( g \) is not always possible for all functions \( f \).
When Does the Inverse Exist?
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Failure in Many-One Functions:
Suppose \( f \) is a many-one function, meaning \( f(a_1) = f(a_2) = b \) for some \( a_1, a_2 \in A \). Then, to define \( g(b) \), we face a problem: should \( g(b) = a_1 \) or \( g(b) = a_2 \)? A function cannot assign multiple outputs to the same input, so \( g \) would fail to be a function in this case. \( g \) would only be a relation from \( B \) to \( A \), not a valid function.
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Failure in Into Functions:
If \( f \) is an into function, then some elements in \( B \) have no preimages in \( A \). For example, if \( b \in B \) has no \( a \in A \) such that \( f(a) = b \), then \( g(b) \) would have no value to assign, violating the definition of a function.
Thus, for \( g \) to exist as a valid function, \( f \) must satisfy two conditions:
- \( f \) must be one-to-one (injective) so that \( g(b) \) is well-defined (no ambiguity in assigning preimages).
- \( f \) must be onto (surjective) so that every \( b \in B \) has a corresponding \( a \in A \).
If \( f \) satisfies these two properties, then \( f \) is said to be bijective, and we call \( f \) invertible.
Definition of the Inverse
Formally, the inverse of \( f : A \to B \), denoted by \( f^{-1} \), is defined as:
If \( f \) is bijective, then \( f^{-1} \) is a function from \( B \to A \). For each \( b \in B \), \( f^{-1}(b) \) gives the unique \( a \in A \) such that \( f(a) = b \).
We can also define inverse like this:
Let \( f : A \to B \) be a function, where \( A \) is the domain and \( B \) is the codomain of \( f \). The function \( f \) is said to be invertible if there exists a function \( g : B \to A \) such that:
- \( g(f(x)) = x \), for all \( x \in A \), and
- \( f(g(y)) = y \), for all \( y \in B \).
If such a function \( g \) exists, it is called the inverse of \( f \) and is denoted by \( f^{-1} : B \to A \).
Furthermore, if \( f \) is invertible, then the inverse \( f^{-1} \) is unique, meaning there is exactly one function \( g \) satisfying the above properties.
Finding inverse
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If \( f \) is given as a set of ordered pairs:
The inverse is obtained by interchanging the components of each ordered pair, which corresponds to interchanging the input and output of the function. Specifically, for each ordered pair \( (a, b) \) in \( f \), the inverse \( f^{-1} \) will contain the pair \( (b, a) \). This highlights the fact that the inverse "reverses" the roles of inputs and outputs.
Let:
\[ f = \{(1, 2), (2, 3), (3, 4)\}. \]Here, the input-output relationship of \( f \) is:
- \( f(1) = 2 \),
- \( f(2) = 3 \),
- \( f(3) = 4 \).
To find the inverse \( f^{-1} \), we reverse these relationships, swapping inputs and outputs. Thus:
\[ f^{-1} = \{(2, 1), (3, 2), (4, 3)\}. \]This means:
- \( f^{-1}(2) = 1 \),
- \( f^{-1}(3) = 2 \),
- \( f^{-1}(4) = 3 \).
The inverse reflects the concept of interchanging input and output values.
- If \( f \) is defined by a formula:
To find the inverse, we interchange \( x \) and \( y \) in the formula \( y = f(x) \). This step corresponds to interchanging the input (\( x \)) and output (\( y \)) of the function. After interchanging, we solve for \( y \) in terms of \( x \), which gives the formula for the inverse function.
Let \( f(x) = 2x + 3 \). Here, the input-output relationship is defined as:
- For an input \( x \), the output is \( f(x) = 2x + 3 \).
To find the inverse \( f^{-1}(x) \), we follow these steps:
- Replace \( f(x) \) with \( y \) for clarity:
\[ y = 2x + 3. \]- Interchange \( x \) and \( y \) to reflect the reversal of inputs and outputs:
\[ x = 2y + 3. \]- Solve for \( y \) in terms of \( x \):
\[ 2y = x - 3 \implies y = \frac{x - 3}{2}. \]Thus, the inverse function is:
\[ f^{-1}(x) = \frac{x - 3}{2}. \]This means:
- If the output of \( f \) is \( x \), the corresponding input is given by \( f^{-1}(x) \).
Inverse of a Quadratic Function
Let \( f : X \to Y \) such that \( f(x) = ax^2 + bx + c \), where \( a > 0 \).
The graph of the quadratic function is a parabola. The vertex of the parabola is
where \( \Delta = b^2 - 4ac \), and its axis is \( x = -\frac{b}{2a} \). If the domain of \( f \), i.e., \( X \), lies in the region left to the axis (or the region right to the axis), then \( f \) is one-to-one, and the codomain \( Y \) is equal to the range.
Conversion of \( f(x) \) into Vertex-Centric Form
Given \( y = ax^2 + bx + c \), start by rewriting \( y \) in a simplified form:
Complete the square inside the parentheses:
This simplifies to:
Now simplify the constant terms:
where \( \Delta = b^2 - 4ac \).
Thus:
Solving for \( x \)
Rearrange the equation to isolate \( \left(x + \frac{b}{2a}\right)^2 \):
Divide through by \( a \):
Take the square root on both sides:
Finally:
Ambiguity of the Sign
The \( \pm \) introduces an ambiguity: which sign to choose? Both signs cannot simultaneously apply, as this would violate the definition of a function (since a function maps one input to exactly one output). Thus, the choice of the sign depends on the domain of \( f(x) \):
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Domain Left of the Axis:
If \( X \) lies to the left of the axis \( x = -\frac{b}{2a} \), choose the negative sign (\( - \)):
\[ x + \frac{b}{2a} = -\sqrt{\frac{y}{a} + \frac{\Delta}{4a^2}}. \]Rearranging:
\[ x = -\frac{b}{2a} - \sqrt{\frac{y}{a} + \frac{\Delta}{4a^2}}. \]Thus:
\[ f^{-1}(x) = -\frac{b}{2a} - \sqrt{\frac{x}{a} + \frac{\Delta}{4a^2}}. \]
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Domain Right of the Axis:
If \( X \) lies to the right of the axis \( x = -\frac{b}{2a} \), choose the positive sign (\( + \)):
\[ x + \frac{b}{2a} = \sqrt{\frac{y}{a} + \frac{\Delta}{4a^2}}. \]Rearranging:
\[ x = -\frac{b}{2a} + \sqrt{\frac{y}{a} + \frac{\Delta}{4a^2}}. \]Thus:
\[ f^{-1}(x) = -\frac{b}{2a} + \sqrt{\frac{x}{a} + \frac{\Delta}{4a^2}}. \]
The ambiguity of the \( \pm \) sign while calculating the inverse occurs because \( f(x) \) is not one-to-one over its entire domain. By restricting the domain to either the left or the right branch of the parabola, \( f(x) \) becomes one-to-one, and the corresponding branch of the inverse function is well-defined. This highlights that the inverse function exists only on specific branches of the domain.
Example
Find the inverse of the function \( f : [2, \infty) \to [-5, \infty) \), where \( f(x) = x^2 - 4x - 1 \).
Solution:
Let \( y = f(x) = x^2 - 4x - 1 \). Rewrite \( y \) by completing the square:
Thus, \( y = (x - 2)^2 - 5 \). Rearranging gives:
Taking the square root of both sides:
Since the domain of \( f(x) \) is \( [2, \infty) \), we have \( x - 2 \geq 0 \), so:
Adding 2 to both sides:
Interchange \( x \) and \( y \) to express the inverse function:
Answer:
The inverse of \( f(x) = x^2 - 4x - 1 \) is:
Example
Find the inverse of the function \( f : (-\infty, -3) \to (-7, \infty) \), where \( f(x) = x^2 + 6x + 2 \).
Solution:
Let \( y = f(x) = x^2 + 6x + 2 \). Rewrite \( y \) by completing the square:
Thus, \( y = (x + 3)^2 - 7 \). Rearranging gives:
Taking the square root of both sides:
Since the domain of \( f(x) \) is \( (-\infty, -3) \), we have \( x + 3 \leq 0 \), so the negative sign is chosen:
Rearranging for \( x \):
Finally, interchange \( x \) and \( y \) to express the inverse function:
Answer:
The inverse of \( f(x) = x^2 + 6x + 2 \) is:
Example
Find the inverse of \( f(x) = e^x - 1 \), where \( x \in \mathbb{R} \).
Solution:
Note: There is no codomain mentioned in the problem. For simplicity, we can assume that the range of the function, which is \( (-1, \infty) \) for \( x \in \mathbb{R} \), is equal to the codomain.
Let \( y = f(x) = e^x - 1 \). Rearrange to solve for \( x \) in terms of \( y \):
Add 1 to both sides:
Take the natural logarithm of both sides:
Interchange \( x \) and \( y \) to express the inverse function:
Answer:
The inverse of \( f(x) = e^x - 1 \) is:
Example
Find the inverse of \( f(x) = e^x - e^{-x} \), where \( x \in \mathbb{R} \).
Solution:
Let \( y = f(x) = e^x - e^{-x} \). Our goal is to isolate \( x \).
From the given \( y = e^x - e^{-x} \) … (1),
we note the identity:
Substituting \( y = e^x - e^{-x} \), we get:
Taking the square root of both sides:
Since \( e^x + e^{-x} > 0 \) for all \( x \in \mathbb{R} \), we choose the positive sign:
Now, add equations (1) and (2):
This simplifies to:
Dividing through by 2:
Take the natural logarithm of both sides to isolate \( x \):
Finally, interchange \( x \) and \( y \) to express the inverse function:
Answer:
The inverse of \( f(x) = e^x - e^{-x} \) is:
Example
Find the inverse of \( f(x) = 2^x + 2^{-x} \), where \( x \geq 0 \).
Solution:
Let \( y = f(x) = 2^x + 2^{-x} \). Our goal is to isolate \( x \).
We note the identity:
Substituting \( y = 2^x + 2^{-x} \), we have:
This implies:
Since \( x \geq 0 \), \( 2^x - 2^{-x} \geq 0 \), so we choose the positive sign:
Now add \( 2^x + 2^{-x} = y \) and \( 2^x - 2^{-x} = \sqrt{y^2 - 4} \):
This simplifies to:
Divide by 2:
Take the logarithm (base 2) on both sides to isolate \( x \):
Finally, interchange \( x \) and \( y \) to express the inverse function:
Answer:
The inverse of \( f(x) = 2^x + 2^{-x} \) is:
Non-Elementary Inverses
In mathematics, it is not always possible to express the inverse of a function in terms of elementary functions (i.e., functions involving polynomials, rational expressions, exponentials, logarithms, trigonometric functions, and their combinations).
For example, consider the function \( f(x) = x + \sin x \), where \( x \in \mathbb{R} \). This function is well-behaved—it is bijective, that can be easily shown. However, the inverse of \( f(x) \) cannot be expressed in terms of elementary functions.
Formal Representation of the Inverse
Let \( y = f(x) = x + \sin x \). To find the inverse, we must solve for \( x \) in terms of \( y \), i.e., solve the equation:
Rewriting:
This equation involves the transcendental function \( \sin x \), and there is no algebraic or elementary manipulation that allows us to isolate \( x \) explicitly in terms of \( y \). Thus, while \( f(x) \) is invertible, its inverse function cannot be written explicitly using elementary functions.
Implicit Definition of the Inverse
Although the inverse cannot be expressed explicitly, it can be implicitly defined. For any \( x \in \mathbb{R} \), the value of \( y = f^{-1}(x) \) satisfies:
This defines \( f^{-1}(x) \) implicitly as the unique value of \( y \) such that the equation holds. The uniqueness is guaranteed because \( f(x) = x + \sin x \) is strictly increasing, ensuring that each \( x \in \mathbb{R} \) corresponds to a single \( y \).
Properties of Inverse Functions
Let \( f \) be a bijective function, and let \( f^{-1} \) denote its inverse. The following are detailed and rigorous properties of inverse functions:
Property I: \( (f^{-1})^{-1} = f \)
Let \( f : A \to B \) be a bijective function, and let \( f^{-1} : B \to A \) be its inverse. We aim to prove that the inverse of \( f^{-1} \) is \( f \), i.e.,
Property II:
Let \( f : A \to B \) and \( g : B \to C \) be two functions. If \( f \) is invertible and \( g \) is invertible, then the composition \( g \circ f : A \to C \) is also invertible. Furthermore, the inverse of \( g \circ f \) is given by:
Proof:
To prove that \( g \circ f : A \to C \) is invertible, we need to show that:
- \( g \circ f \) is one-to-one (injective),
- \( g \circ f \) is onto (surjective),
- The inverse of \( g \circ f \) satisfies \( (g \circ f)^{-1} = f^{-1} \circ g^{-1} \).
One-to-One
To prove that \( g \circ f \) is one-to-one, let \( x_1, x_2 \in A \) such that:
Since \( g \) is invertible, it is one-to-one. Therefore, we have:
Similarly, since \( f \) is also invertible and one-to-one, it follows that:
Thus, \( g \circ f \) is one-to-one.
Onto
To prove that \( g \circ f \) is onto, let \( c \in C \) be an arbitrary element. Since \( g \) is invertible and onto, there exists some \( b \in B \) such that:
Further, since \( f \) is invertible and onto, there exists some \( a \in A \) such that:
Substituting \( b = f(a) \) into \( g(b) = c \), we have:
Therefore, \( g \circ f(a) = c \), which proves that \( g \circ f \) is onto.
Inverse of \( g \circ f \)
To construct the inverse of \( g \circ f \), note that:
- \( g \) has an inverse \( g^{-1} : C \to B \), and
- \( f \) has an inverse \( f^{-1} : B \to A \).
For \( c \in C \), let \( b = g^{-1}(c) \), so \( g(b) = c \). Since \( b \in B \), let \( a = f^{-1}(b) \), so \( f(a) = b \). Combining these, we have:
Thus, the inverse of \( g \circ f \) is:
The composition \( g \circ f : A \to C \) is invertible, and its inverse is given by:
This result shows that the invertibility of \( f \) and \( g \) guarantees the invertibility of their composition, with the inverse following the order of composition in reverse.
Graphical Relationship between a Function and its inverse
Consider a point \( (a, b) \) in the Cartesian plane. Its reflection in the line \( y = x \) can be obtained using the formula for the image of a point in a line. Using the concepts of coordinate geometry, the image of \( (a, b) \) in \( y = x \) is found by interchanging the coordinates. Therefore, the reflection of \( (a, b) \) in \( y = x \) is \( (b, a) \).
Now, consider a function \( y = f(x) \). By definition of the inverse function, we have:
This means the point \( (x, y) \) lies on the graph of \( f \), and the point \( (y, x) \) lies on the graph of \( f^{-1} \).
Clearly, the point \( (y, x) \) is the reflection of \( (x, y) \) across the line \( y = x \). Since this relationship holds for every \( (x, y) \) on the graph of \( f \), it follows that every point on the graph of \( f \) has a corresponding image on the graph of \( f^{-1} \), reflected across \( y = x \).
Thus, the graphs of \( f \) and \( f^{-1} \) are symmetric about the line \( y = x \). The reflection property ensures that for every \( (a, b) \) on \( f \), the corresponding point \( (b, a) \) lies on \( f^{-1} \). This symmetry is fundamental to understanding the relationship between a function and its inverse.
Example
The graph illustrates the geometric relationship between a function \( f(x) = x^3 + x - 2 \) and its inverse \( f^{-1}(x) \). The line \( y = x \) acts as the axis of symmetry, reflecting every point on \( f(x) \) to its corresponding point on \( f^{-1}(x) \).
- The blue curve represents \( f(x) = x^3 + x - 2 \).
- The red dashed curve shows \( f^{-1}(x) \), obtained by swapping \( x \) and \( y \) coordinates of \( f(x) \).
- The green dotted line (\( y = x \)) demonstrates the symmetry between \( f(x) \) and \( f^{-1}(x) \). Highlighted points and connecting lines visually confirm the reflection property, where \( (x, y) \) on \( f(x) \) maps to \( (y, x) \) on \( f^{-1}(x) \).
Example
Geometrically, the functions \( a^x \) and \( \log_a x \) are reflections of each other across the line \( y = x \).
- For \( a^x \), a point \( (x, a^x) \) lies on its graph.
- For \( \log_a x \), the point \( (a^x, x) \) lies on its graph.
Thus, each point \( (x, a^x) \) on \( a^x \) has a corresponding point \( (a^x, x) \) on \( \log_a x \), and this reflection across \( y = x \) demonstrates their inverse relationship graphically.
The graph above illustrates the functions \( f(x) = 2^x \) (blue curve) and \( f^{-1}(x) = \log_2(x) \) (red dashed curve), along with the line \( y = x \) (green dotted line).
The reflection property is clearly shown: every point \( (x, 2^x) \) on \( f(x) \) corresponds to \( (2^x, x) \) on \( f^{-1}(x) \), demonstrating their symmetry across the line \( y = x \). Specific points and their reflections are highlighted for clarity.
Intersection of \( f \) and \( f^{-1} \) Based on Their increasing and decreasing Behavior
For \( f \) to be invertible, it must be one-one (injective). A one-one function implies that \( f \) must be either always increasing (strictly increasing) or always decreasing (strictly decreasing).
If \( f \) is strictly increasing and \( f \) intersects its inverse \( f^{-1} \), the points of intersection will always lie on the line \( y = x \).
- For strictly increasing functions, \( f(x) = f^{-1}(x) \) implies \( f(x) = x \).
- This happens because, geometrically, \( f \) and \( f^{-1} \) are reflections of each other across the line \( y = x \). Thus, any intersection point must satisfy \( x = y \), lying exactly on \( y = x \).
Example:
For \( f(x) = x^3 \), which is strictly increasing, its inverse \( f^{-1}(x) = \sqrt[3]{x} \) intersects at points satisfying \( x^3 = x \). Solving this gives intersection points at \( x = -1, 0, 1 \), all of which lie on \( y = x \).
But, if \( f \) is strictly decreasing and intersects its inverse \( f^{-1} \), the points of intersection may not lie on the line \( y = x \).
- For strictly decreasing functions, \( f(x) = f^{-1}(x) \) no longer guarantees that \( x = y \). This is because the reflection property across \( y = x \) does not align with the behavior of a decreasing function.
Example:
For \( f(x) = -x^3 \), a strictly decreasing function, its inverse \( f^{-1}(x) = -\sqrt[3]{x} \) intersects at points satisfying \( -x^3 = -x \), or \( x^3 = x \). Solving this gives \( x = -1, 0, 1 \). However, while \( x = 0 \) lies on \( y = x \), the other intersections may not align with this line due to the decreasing nature of \( f \).
Therefore, for strictly increasing functions, the intersection points of \( f \) and \( f^{-1} \) always lie on \( y = x \). For strictly decreasing functions, the intersection points may or may not lie on \( y = x \), depending on the specific function.
Example
Solve the equation:
Solution:
Let \( y = 2x^2 - 4x - 3 \). The given equation becomes:
Step 1: Isolate \( x \) on the Right-Hand Side
Rearranging the equation for \( y \):
Squaring both sides:
Multiply through by 2:
Rearranging for \( x \):
Thus, the right-hand side \( 1 + \sqrt{\frac{x + 5}{2}} \) is the inverse of the left-hand side \( y = 2x^2 - 4x - 3 \).
Step 2: Behavior of \( f(x) = 2x^2 - 4x - 3 \) for \( x \geq 1 \)
The function \( f(x) = 2x^2 - 4x - 3 \) is strictly increasing for \( x \geq 1 \) (you can chack that using graph or simply derivatives)
Since \( f(x) \) is strictly increasing, the intersection point of \( f(x) \) and \( f^{-1}(x) \) must lie on the line \( y = x \). Therefore, we solve:
Step 3: Solve \( 2x^2 - 4x - 3 = x \)
Rearranging the equation:
Factoring or using the quadratic formula:
This gives:
Step 4: Check for Validity
Since \( x \geq 1 \), only \( x = 3 \) is valid.
The solution to the equation is: