How Do You Know If A Function Is Invertible

Invertibility of a function is a fundamental concept in mathematics, particularly in fields like calculus, algebra, and analysis. A function is invertible if there exists another function that can "undo" its effect. Understanding how to determine if a function is invertible is essential for solving equations, simplifying expressions, and gaining deeper insights into mathematical relationships.

What is an Invertible Function?

An invertible function, also known as a bijective function, is a function where each element of the range is associated with exactly one element of the domain. In simpler terms, a function f(x) is invertible if there exists another function g(x) such that:

• g(f(x)) = x for all x in the domain of f
• f(g(x)) = x for all x in the domain of g

The function g(x) is called the inverse of f(x), often denoted as f⁻¹(x).

Key Properties of Invertible Functions

• One-to-One (Injective): Each element in the domain maps to a unique element in the range. No two elements in the domain map to the same element in the range.
• Onto (Surjective): Every element in the range has a corresponding element in the domain. The range of the function is equal to its codomain.
• Bijective: A function that is both one-to-one (injective) and onto (surjective). Only bijective functions are invertible.

Methods to Determine Invertibility

Several methods can be used to determine whether a function is invertible. These methods include graphical tests, algebraic methods, and calculus-based approaches.

1. Horizontal Line Test (Graphical Method)

The Horizontal Line Test is a simple graphical method to determine if a function is one-to-one, which is a prerequisite for invertibility.

How it Works:

1. Graph the function: Plot the function f(x) on a coordinate plane.
2. Draw horizontal lines: Imagine drawing horizontal lines across the graph.
3. Check for intersections: If any horizontal line intersects the graph more than once, the function is not one-to-one and, therefore, not invertible. If every horizontal line intersects the graph at most once, the function is one-to-one and may be invertible.

Example:

• Consider the function f(x) = x². If you graph this function, you'll see that a horizontal line, such as y = 4, intersects the graph at x = 2 and x = -2. This means f(x) = x² is not one-to-one over its entire domain (all real numbers) and is not invertible.
• However, if we restrict the domain of f(x) = x² to x ≥ 0, the function becomes one-to-one. Now, any horizontal line will intersect the graph at most once, making the function invertible over this restricted domain.

2. Algebraic Method

The algebraic method involves manipulating the function's equation to solve for x in terms of y.

How it Works:

1. Replace f(x) with y: Rewrite the function as y = f(x).
2. Solve for x: Rearrange the equation to isolate x in terms of y.
3. Check for uniqueness: If you can solve for x uniquely in terms of y, the function is one-to-one. If there are multiple possible values of x for a given y, the function is not one-to-one and not invertible.
4. Verify the inverse: If you find a unique solution for x, you can write the inverse function as f⁻¹(y) = x. To verify, check that f(f⁻¹(y)) = y and f⁻¹(f(x)) = x.

Examples:

• Example 1: Linear Function

  Consider the function f(x) = 2x + 3.

  1. Replace f(x) with y: y = 2x + 3
  2. Solve for x:
     • y - 3 = 2x
     • x = (y - 3) / 2
  3. Since we have a unique solution for x in terms of y, the function is one-to-one.
  4. The inverse function is f⁻¹(y) = (y - 3) / 2.
  5. Verification:
     • f(f⁻¹(y)) = 2((y - 3) / 2) + 3 = (y - 3) + 3 = y
     • f⁻¹(f(x)) = ((2x + 3) - 3) / 2 = (2x) / 2 = x

  Thus, f(x) = 2x + 3 is invertible, and its inverse is f⁻¹(x) = (x - 3) / 2.

• Example 2: Quadratic Function

  Consider the function f(x) = x².

  1. Replace f(x) with y: y = x²
  2. Solve for x:
     • x = ±√y
  3. Since we have two possible values for x (positive and negative square roots), the function is not one-to-one over its entire domain (all real numbers). Therefore, f(x) = x² is not invertible without restricting its domain.

• Example 3: Rational Function

  Consider the function f(x) = (x + 1) / (x - 1).

  1. Replace f(x) with y: y = (x + 1) / (x - 1)
  2. Solve for x:
     • y(x - 1) = x + 1
     • yx - y = x + 1
     • yx - x = y + 1
     • x(y - 1) = y + 1
     • x = (y + 1) / (y - 1)
  3. Since we have a unique solution for x in terms of y, the function is one-to-one.
  4. The inverse function is f⁻¹(y) = (y + 1) / (y - 1). Notice that the inverse function is the same as the original function in this case.
  5. Verification:
     • f(f⁻¹(y)) = (((y + 1) / (y - 1)) + 1) / (((y + 1) / (y - 1)) - 1) = (((y + 1) + (y - 1)) / (y - 1)) / (((y + 1) - (y - 1)) / (y - 1)) = (2y / (y - 1)) / (2 / (y - 1)) = y
     • f⁻¹(f(x)) = ((x + 1) / (x - 1) + 1) / ((x + 1) / (x - 1) - 1) = (((x + 1) + (x - 1)) / (x - 1)) / (((x + 1) - (x - 1)) / (x - 1)) = (2x / (x - 1)) / (2 / (x - 1)) = x

  Thus, f(x) = (x + 1) / (x - 1) is invertible, and its inverse is f⁻¹(x) = (x + 1) / (x - 1).

3. Calculus-Based Methods

Calculus provides tools to determine the invertibility of a function, particularly through the use of derivatives.

a. Derivative Test

The derivative of a function can help determine if the function is strictly increasing or strictly decreasing, which implies it is one-to-one and therefore invertible.

How it Works:

1. Find the derivative: Calculate the derivative f'(x) of the function f(x).
2. Analyze the sign of the derivative:
   • If f'(x) > 0 for all x in the domain, the function is strictly increasing and therefore one-to-one.
   • If f'(x) < 0 for all x in the domain, the function is strictly decreasing and therefore one-to-one.
   • If f'(x) ≥ 0 or f'(x) ≤ 0 but f'(x) is not identically zero on any interval, the function is monotonic and one-to-one.
   • If f'(x) changes sign, the function is not strictly increasing or decreasing and may not be one-to-one.
3. Check for invertibility: If the function is strictly increasing or strictly decreasing, it is one-to-one and invertible.

Examples:

• Example 1: f(x) = x³

  1. Find the derivative: f'(x) = 3x²
  2. Analyze the sign of the derivative: f'(x) = 3x² ≥ 0 for all x. The derivative is zero only at x = 0, but it does not change sign. Therefore, the function is strictly increasing.
  3. Check for invertibility: Since f(x) = x³ is strictly increasing, it is one-to-one and invertible. Its inverse is f⁻¹(x) = ∛x.

• Example 2: f(x) = x²

  1. Find the derivative: f'(x) = 2x
  2. Analyze the sign of the derivative: f'(x) = 2x is positive for x > 0 and negative for x < 0. The derivative changes sign, indicating that the function is not strictly increasing or decreasing over its entire domain.
  3. Check for invertibility: Since f(x) = x² is not strictly increasing or decreasing, it is not one-to-one over its entire domain and is not invertible without restricting its domain.

• Example 3: f(x) = eˣ

  1. Find the derivative: f'(x) = eˣ
  2. Analyze the sign of the derivative: f'(x) = eˣ > 0 for all x. Therefore, the function is strictly increasing.
  3. Check for invertibility: Since f(x) = eˣ is strictly increasing, it is one-to-one and invertible. Its inverse is f⁻¹(x) = ln(x), defined for x > 0.

b. Second Derivative Test (for Concavity)

While the second derivative test is primarily used to find local maxima and minima, it can provide additional information about the function's behavior that may help in determining invertibility, especially in conjunction with the first derivative test.

How it Works:

1. Find the first derivative: Calculate the first derivative f'(x) of the function f(x).
2. Find the second derivative: Calculate the second derivative f''(x) of the function f(x).
3. Analyze the concavity:
   • If f''(x) > 0 for all x in the domain, the function is concave up.
   • If f''(x) < 0 for all x in the domain, the function is concave down.
4. Combine with the first derivative test: Use the information about concavity along with the sign of the first derivative to determine if the function is strictly increasing or decreasing.

Example:

• Consider the function f(x) = x³.

  1. Find the first derivative: f'(x) = 3x²
  2. Find the second derivative: f''(x) = 6x
  3. Analyze the concavity: f''(x) = 6x is positive for x > 0 (concave up) and negative for x < 0 (concave down). Although the concavity changes, the first derivative is always non-negative, indicating that the function is strictly increasing and therefore invertible.

4. Composition Test

The composition test is a direct method based on the definition of an inverse function.

How it Works:

1. Guess a potential inverse: Suppose you have a function f(x) and you suspect that g(x) is its inverse.
2. Compute the compositions:
   • Calculate f(g(x))
   • Calculate g(f(x))
3. Check if the compositions equal x:
   • If f(g(x)) = x for all x in the domain of g and g(f(x)) = x for all x in the domain of f, then g(x) is the inverse of f(x) and f(x) is invertible.
   • If either composition does not equal x, then g(x) is not the inverse of f(x) and you need to find another potential inverse or conclude that f(x) is not invertible.

Examples:

• Example 1: f(x) = 2x + 3 and g(x) = (x - 3) / 2

  1. Compute f(g(x)):
     • f(g(x)) = 2((x - 3) / 2) + 3 = (x - 3) + 3 = x
  2. Compute g(f(x)):
     • g(f(x)) = ((2x + 3) - 3) / 2 = (2x) / 2 = x
  3. Since both compositions equal x, g(x) = (x - 3) / 2 is the inverse of f(x) = 2x + 3, and f(x) is invertible.

• Example 2: f(x) = x² and g(x) = √x

  1. Compute f(g(x)):
     • f(g(x)) = (√x)² = x for x ≥ 0
  2. Compute g(f(x)):
     • g(f(x)) = √(x²) = |x|
  3. Since g(f(x)) = |x| is not equal to x for all x (it only equals x for x ≥ 0), g(x) = √x is not the inverse of f(x) = x² over the entire domain of real numbers. However, if we restrict the domain of f(x) to x ≥ 0, then g(x) = √x is its inverse.

Practical Considerations and Common Functions

Restricting the Domain

As seen with the example of f(x) = x², a function that is not one-to-one over its entire domain can often be made invertible by restricting the domain. This is a common technique used with trigonometric functions and other functions that exhibit periodic or symmetrical behavior.

• Example: f(x) = sin(x)

  The sine function is not one-to-one over its entire domain (all real numbers) because it oscillates between -1 and 1. However, if we restrict the domain to [-π/2, π/2], the sine function becomes strictly increasing and therefore one-to-one. The inverse function is f⁻¹(x) = arcsin(x) or sin⁻¹(x), defined for x in the range [-1, 1].

Common Invertible Functions

• Linear Functions: f(x) = ax + b, where a ≠ 0, are always invertible.
• Exponential Functions: f(x) = aˣ, where a > 0 and a ≠ 1, are invertible. Their inverses are logarithmic functions.
• Logarithmic Functions: f(x) = logₐ(x), where a > 0 and a ≠ 1, are invertible. Their inverses are exponential functions.
• Odd Power Functions: f(x) = xⁿ, where n is an odd integer, are invertible.
• Certain Rational Functions: Some rational functions, like f(x) = (x + 1) / (x - 1), are invertible.

Functions That Are Not Invertible

• Even Power Functions: f(x) = xⁿ, where n is an even integer, are not invertible over their entire domain (all real numbers) without restriction.
• Constant Functions: f(x) = c, where c is a constant, are not invertible because they map every element in the domain to the same element in the range, violating the one-to-one property.
• Periodic Functions: Functions like sin(x), cos(x), and tan(x) are not invertible over their entire domain without restriction.

Conclusion

Determining whether a function is invertible involves checking if it is both one-to-one (injective) and onto (surjective), which together define a bijective function. Methods such as the Horizontal Line Test, algebraic manipulation, and calculus-based approaches, including the derivative test, provide effective ways to assess invertibility. By understanding these methods and considering practical aspects like domain restriction, one can confidently determine if a function has an inverse and, if so, find its inverse function. Invertible functions play a crucial role in many areas of mathematics and are essential for solving equations and understanding mathematical relationships.