Inverse Of A Function Examples With Solutions

Inverse of a Function Examples with Solutions

Inverse functions are a fundamental concept in mathematics that help us "reverse" the effect of a given function. Even so, the inverse function, if it exists, takes that output and returns us back to the original input. Practically speaking, when we apply a function to an input, we get an output. Now, understanding how to find and work with inverse functions is essential in various branches of mathematics, from algebra to calculus and beyond. In this article, we'll explore the concept of inverse functions through detailed examples with step-by-step solutions.

Understanding Inverse Functions

An inverse function essentially undoes what the original function does. If we have a function f(x) that maps x to y, then its inverse function, denoted as f^(-1)(x), maps y back to x. For a function to have an inverse, it must be bijective, meaning it must be both injective (one-to-one) and surjective (onto). In simpler terms, each input must correspond to exactly one output, and each output must correspond to exactly one input.

Not all functions have inverses. That's why for example, the function f(x) = x² does not have an inverse over its entire domain because both positive and negative inputs (like 2 and -2) produce the same output (4). Still, if we restrict the domain to non-negative numbers, then f(x) = x² does have an inverse.

Finding the Inverse of a Function: Step-by-Step Process

To find the inverse of a function, follow these steps:

1. Verify that the function is one-to-one (passes the horizontal line test)
2. Replace f(x) with y
3. Swap x and y in the equation
4. Solve for y
5. Replace y with f^(-1)(x)

Let's apply this process to several examples with detailed solutions.

Examples with Solutions

Example 1: Linear Function

Find the inverse of f(x) = 3x + 5.

Solution:

1. First, we verify that this linear function is one-to-one. Since it's a straight line with a non-zero slope, it passes the horizontal line test.
2. Replace f(x) with y: y = 3x + 5
3. Swap x and y: x = 3y + 5
4. Solve for y: x - 5 = 3y y = (x - 5)/3
5. Replace y with f^(-1)(x): f^(-1)(x) = (x - 5)/3

Because of this, the inverse of f(x) = 3x + 5 is f^(-1)(x) = (x - 5)/3.

Example 2: Quadratic Function with Restricted Domain

Find the inverse of f(x) = x² for x ≥ 0 And that's really what it comes down to..

Solution:

1. The function f(x) = x² is not one-to-one over all real numbers, but when we restrict the domain to x ≥ 0, it becomes one-to-one.
2. Replace f(x) with y: y = x²
3. Swap x and y: x = y²
4. Solve for y: Since we've restricted the domain to x ≥ 0, we only take the non-negative square root: y = √x
5. Replace y with f^(-1)(x): f^(-1)(x) = √x

Because of this, the inverse of f(x) = x² for x ≥ 0 is f^(-1)(x) = √x.

Example 3: Rational Function

Find the inverse of f(x) = (2x + 3)/(x - 1) It's one of those things that adds up..

Solution:

1. First, we need to verify that this rational function is one-to-one. A rational function like this is typically one-to-one as long as the denominator doesn't cause any issues with horizontal lines intersecting the graph more than once.
2. Replace f(x) with y: y = (2x + 3)/(x - 1)
3. Swap x and y: x = (2y + 3)/(y - 1)
4. Solve for y: Multiply both sides by (y - 1): x(y - 1) = 2y + 3 Distribute x: xy - x = 2y + 3 Get all terms with y on one side: xy - 2y = x + 3 Factor out y: y(x - 2) = x + 3 Solve for y: y = (x + 3)/(x - 2)
5. Replace y with f^(-1)(x): f^(-1)(x) = (x + 3)/(x - 2)

So, the inverse of f(x) = (2x + 3)/(x - 1) is f^(-1)(x) = (x + 3)/(x - 2).

Example 4: Exponential Function

Find the inverse of f(x) = e^(2x + 1).

Solution:

1. Exponential functions with non-zero exponents are one-to-one.
2. Replace f(x) with y: y = e^(2x + 1)
3. Swap x and y: x = e^(2y + 1)
4. Solve for y: Take the natural logarithm of both sides: ln(x) = 2y + 1 Subtract 1: ln(x) - 1 = 2y Divide by 2: y = (ln(x) - 1)/2
5. Replace y with f^(-1)(x): f^(-1)(x) = (ln(x) - 1)/2

Because of this, the inverse of f(x) = e^(2x + 1) is f^(-1)(x) = (ln(x) - 1)/2 Worth knowing..

Example 5: Trigonometric Function

Find the inverse of f(x) = sin(x) for -π/2 ≤ x ≤ π/2 Easy to understand, harder to ignore..

Solution:

1. The sine function is periodic and not one-to-one over its entire domain, but when we restrict it to -π/2 ≤ x ≤ π/2, it becomes one-to-one.
2. Replace f

(x) with y: y = sin(x) 3. Swap x and y: x = sin(y) 4. Solve for y: To isolate y, apply the inverse sine function (arcsin) to both sides: y = arcsin(x) 5 The details matter here..

That's why, the inverse of f(x) = sin(x) for the restricted domain is f⁻¹(x) = arcsin(x).

Summary Table of Inverse Operations

To quickly identify how to solve for an inverse, it is helpful to remember the relationship between standard functions and their inverses:

Some disagree here. Fair enough.

Conclusion

Finding the inverse of a function is a systematic process that involves switching the roles of the independent and dependent variables. Whether you are working with simple linear equations, complex rational functions, or transcendental functions like exponentials and trigonometry, the core steps remain the same: replace the function notation with $y$, swap $x$ and $y$, and use algebraic manipulation to isolate the new $y$.

Easier said than done, but still worth knowing.

Crucially, always remember to check if a function is one-to-one. Which means if a function fails the horizontal line test, such as a parabola, you must restrict its domain to see to it that a unique inverse exists. Mastering these steps allows you to "undo" mathematical operations, a skill that is fundamental to higher-level calculus and algebraic analysis.