What is the inverse of the function shown is a common question when students encounter a graph, table, or algebraic expression and need to determine the rule that reverses the original mapping. Understanding how to find an inverse function is essential for solving equations, modeling real‑world phenomena, and progressing to higher‑level mathematics such as calculus. This article walks through the concept, provides a step‑by‑step method, illustrates the process with several typical functions, and clarifies frequent pitfalls.
Introduction
When a function f takes an input x and produces an output y (written y = f(x)), its inverse f⁻¹ does the opposite: it takes the output y and returns the original input x. In notation, if y = f(x), then x = f⁻¹(y), or equivalently f⁻¹(f(x)) = x and f(f⁻¹(y)) = y for all values in the appropriate domains.
The phrase what is the inverse of the function shown often appears in textbooks where a function is presented graphically or in a table. So the goal is to derive the algebraic expression of f⁻¹ and, if necessary, state its domain and range. The following sections break down the procedure into clear, actionable steps.
Understanding Functions and Their Inverses
One‑to‑One Requirement
A function possesses an inverse only if it is one‑to‑one (injective). This means each y value comes from exactly one x value. Now, graphically, a function passes the horizontal line test: no horizontal line intersects the curve more than once. If the original function fails this test, its inverse will not be a function unless we restrict the domain Took long enough..
Domain and Range Swap
For a valid pair f and f⁻¹:
- The domain of f becomes the range of f⁻¹.
- The range of f becomes the domain of f⁻¹.
Keeping track of this swap prevents errors when stating the final answer.
Steps to Find the Inverse of a Function Shown
Whether the function is given as an equation, a table, or a graph, the same logical sequence applies.
-
Write the function in y = form
If the expression is already solved for y, proceed. If not, isolate y on one side (e.g., from 2x + 3y = 6 rewrite as y = (6 – 2x)/3) No workaround needed.. -
Swap the variables
Replace every x with y and every y with x. This step reflects the reversal of input and output It's one of those things that adds up.. -
Solve the new equation for y
After swapping, isolate y again. The resulting expression is f⁻¹(x) It's one of those things that adds up.. -
State the domain and range of the inverse
Use the original function’s range and domain, respectively. If a domain restriction was needed to make the function one‑to‑one, apply the same restriction to the inverse’s domain. -
Verify (optional but recommended)
Compose the functions: check that f(f⁻¹(x)) = x and f⁻¹(f(x)) = x for values within the domains.
These steps work for linear, polynomial, exponential, logarithmic, and trigonometric functions, provided the one‑to‑one condition holds (or is enforced) And that's really what it comes down to..
Examples of Finding the Inverse
Example 1: Linear Function
Suppose the shown function is f(x) = 3x – 5 It's one of those things that adds up..
- Write as y = 3x – 5.
- Swap: x = 3y – 5.
- Solve for y:
- Add 5: x + 5 = 3y
- Divide by 3: y = (x + 5)/3
Hence, f⁻¹(x) = (x + 5)/3.
Domain and range of f are all real numbers (ℝ), so the inverse also has domain ℝ and range ℝ.
Example 2: Quadratic Function with Domain Restriction
Consider the shown function f(x) = x² but only for x ≥ 0 (the right half of the parabola).
- y = x², with x ≥ 0.
- Swap: x = y².
- Solve for y: take the square root. Because the original domain restricted x to non‑negative values, the inverse must output non‑negative numbers: y = √x.
Thus, f⁻¹(x) = √x.
Domain of f: [0, ∞) → becomes range of f⁻¹.
Range of f: [0, ∞) → becomes domain of f⁻¹.
If the restriction had been x ≤ 0, the inverse would be f⁻¹(x) = –√x.
Example 3: Exponential Function
Let the shown function be f(x) = 2ˣ.
- y = 2ˣ.
- Swap: x = 2ʸ.
- Solve for y using logarithms: y = log₂(x).
So, f⁻¹(x) = log₂(x).
Domain of f: ℝ → range of f⁻¹: ℝ.
Range of f: (0, ∞) → domain of f⁻¹: (0, ∞).
Example 4: Trigonometric Function (Sine)
The sine function is not one‑to‑one over its entire period, but the standard restriction [–π/2, π/2] yields an invertible piece.
Shown function: f(x) = sin(x) for x ∈ [–π/2, π/2].
- y = sin(x).
Example 4 continued – the restricted sine function
The given piece of the sine curve is defined on the interval ([-\pi/2,;\pi/2]).
Starting from (y=\sin x) we interchange the roles of the variables, obtaining
(x=\sin y). To isolate (y) we invoke the inverse sine (arcsine) function, which is precisely the operation that undoes the sine on this restricted domain:
[ y=\arcsin x . ]
Thus the inverse function is (f^{-1}(x)=\arcsin x).
Domain and range.
Because the original function’s domain was limited to ([-\pi/2,;\pi/2]), its range becomes the domain of the inverse, namely ([‑1,1]). Conversely, the original range ([‑1,1]) becomes the domain of the inverse, which remains ([‑\pi/2,;\pi/2]).
Verification.
Compose the two expressions:
[ f\bigl(f^{-1}(x)\bigr)=\sin\bigl(\arcsin x\bigr)=x\qquad\text{for }x\in[‑1,1], ]
[ f^{-1}\bigl(f(x)\bigr)=\arcsin\bigl(\sin x\bigr)=x\qquad\text{for }x\in[‑\pi/2,;\pi/2]. ]
Both identities hold, confirming that the derived formula is correct Worth keeping that in mind..
Example 5 – a rational function
Consider the function
[ f(x)=\frac{2x+1}{,x-3,},\qquad x\neq 3 . ]
- Express the relationship as (y=\dfrac{2x+1}{x-3}).
- Swap the variables: (x=\dfrac{2y+1}{y-3}).
- Solve for (y). Multiply both sides by the denominator:
[ x(y-3)=2y+1;\Longrightarrow;xy-3x=2y+1. ]
Gather the terms containing (y) on one side:
[ xy-2y=3x+1;\Longrightarrow;y(x-2)=3x+1. ]
Finally, isolate (y):
[ y=\frac{3x+1}{x-2}. ]
Hence
[ f^{-1}(x)=\frac{3x+1}{x-2},\qquad x\neq 2 . ]
Domain and range.
The original function is undefined at (x=3); its range excludes the value (y=2) (the horizontal asymptote). As a result, the inverse’s domain is all real numbers except (2), and its range excludes (3) And it works..
Verification.
[
f\bigl(f^{-1}(x)\bigr)=\frac{2\left(\frac{3x+1}{x-2}\right)+1}{\frac{3x+1}{x-2}-3}=x,
]
[ f^{-1}\bigl(f(x)\bigr)=\frac{3\left(\frac{2x+1}{x-3}\right)+1}{\frac{2x+1}{x-3}-2}=x, ]
both valid for the appropriate exclusions, confirming the correctness of the inverse.
Example 6 – a logarithmic function
Let
[ f(x)=\ln!\bigl(x+4\bigr)-1,\qquad x>-4 . ]
- Write (y=\ln(x+4)-1).
- Exchange the variables: (x=\ln(y+4)-1).
- Isolate the logarithm: (x+1=\ln(y+4)).
- Exponentiate to remove the log: (e^{,x+1}=y+4).
- Solve for (y):
[ y=e^{,x+1}-4. ]
Thus
[ f^{-1}(x)=e^{,x+1}-4,\qquad x\in\mathbb{R}. ]
Domain and range.
The original domain ((-4,\infty)) becomes the range of the inverse, which is ((‑4,\infty)). The original range ((‑\infty,\infty)) becomes the domain of the inverse, i.e., all real numbers Nothing fancy..
Verification.
[ f\bigl(f^{-1}(x)\bigr)=\ln!\bigl(e^{,x+1}-4+4\bigr)-1=\ln!\bigl(e^{,x+1}\bigr)-1=(x+1)-1=x, ]
[ f^{-1}\bigl(f(x)\bigr)=e^{,\bigl(\ln(x+4)-1\bigr)+1}-4=e^{,\ln(x+4)}-4=(x+4)-4=x, ]
for every admissible (x).
Conclusion
Finding the inverse of a function is a systematic process that hinges on three core ideas: rewriting the relationship so that the dependent variable is isolated, interchanging the roles of the independent and dependent variables, and then solving again for the new dependent variable. When the original function is not one‑to‑one over its entire domain, a suitable restriction must be imposed first; the same restriction then applies to the inverse’s domain.
After constructing the inverse, checking that the compositions yield the identity function confirms that the algebraic work was performed correctly. The domain and range of the inverse are simply the swapped ranges of the original function, reflecting the fact that inputs and outputs exchange places.
Mastering these steps equips the reader to handle a wide variety of functions — linear, polynomial, exponential, logarithmic, trigonometric, and rational — thereby providing a versatile toolkit for solving real‑world problems that require reversing the effect of a mathematical model Simple, but easy to overlook. That alone is useful..
