Inverse Functions

Let’s unravel the world of inverse functions! 🔄✨

What are Inverse Functions? 🤔

Imagine a magical function machine f 🤖 that takes an input x and spits out an output f(x). An inverse function, denoted as f⁻¹ (read as “f inverse”), is like a reverse magic machine 🪄. Its job is to undo what the original f machine did. If you put f(x) into f⁻¹, it will give you back the original x! [5.10]

Formally, for a function f and its inverse f⁻¹, the following compositions hold true [5.10]:

• (f⁻¹ ◦ f)(x) = f⁻¹(f(x)) = x 🔙 (Start with x, apply f, then f⁻¹, and you’re back to x!)
• (f ◦ f⁻¹)(x) = f(f⁻¹(x)) = x 🔁 (Start with x, apply f⁻¹, then f, and you’re also back to x!)

The “One-to-One” Rule: A Crucial Requirement! ☝️

For an inverse function to exist, the original function f must be a one-to-one function [5.10]. What does “one-to-one” mean? 🤔

• A function f is one-to-one (or injective) if each element in its input set (domain) maps to a distinct element in its output set (range) [1.6.1]. In simpler terms, no two different inputs can produce the same output. 🙅‍♀️🙅‍♂️
• Emoji Analogy: Imagine a class where every student has a unique student ID 🆔. If two students had the same ID, you wouldn’t know which student to find if you only had the ID!
• Graphical Test: The Horizontal Line Test 📏➡️📈
  • You can visually check if a function is one-to-one by using the Horizontal Line Test [5.2]. If any horizontal line crosses the graph of the function at most once, then the function is one-to-one [5.2]. If it crosses more than once, it’s not one-to-one, and therefore, it does not have an inverse function.
  • Functions that are consistently increasing or consistently decreasing are always one-to-one [5.2].

Domain and Range Swap: A Magical Exchange! 🌍↔️📦

One of the coolest properties of inverse functions is how their domains and ranges are related [5.10]:

• The domain of f becomes the range of f⁻¹ 🌍 = 📦
• The range of f becomes the domain of f⁻¹ 📦 = 🌍

This makes sense, as the inverse function literally “reverses” the mapping of inputs and outputs! [5.10]

Graphical Symmetry: A Mirror Image! 🖼️

When you graph a function f and its inverse f⁻¹ on the same coordinate plane, you’ll notice a beautiful symmetry. The graph of f⁻¹ is a reflection of the graph of f across the line y = x [5.10].

• Emoji Analogy: Think of the line y = x as a perfect mirror 🪞. If you fold the paper along this line, the graph of f would perfectly overlap with the graph of f⁻¹.
• This also means that if a point (a, f(a)) is on the graph of f, then the point (f(a), a) will be on the graph of f⁻¹ [5.10].

Important Note: Not a Reciprocal! ⚠️

Be careful! The notation f⁻¹(x) does not mean 1/f(x) [5.10]. It’s a special notation for the inverse function.

Examples from the Sources: Confirming the Magic! ✨

The sources demonstrate how to verify if two functions are inverses by checking their composition [5.10].

• Example 1: Linear Functions 📏

  • Given g(x) = 4x and h(x) = x/4. To verify if g is the inverse of h (and vice-versa), we check:
    • (g ◦ h)(x) = g(h(x)) = g(x/4) = 4 * (x/4) = x ✅
    • (h ◦ g)(x) = h(g(x)) = h(4x) = (4x)/4 = x ✅
  • Since both compositions result in x, g(x) and h(x) are indeed inverse functions [5.10].

• Example 2: Cubic Functions 🧊

  • Given g(x) = x³ and g⁻¹(x) = x^(1/3). To verify:
    • g⁻¹(g(x)) = g⁻¹(x³) = (x³)^(1/3) = x ✅
    • g(g⁻¹(x)) = g(x^(1/3)) = (x^(1/3))³ = x ✅
  • Thus, they are inverses [5.10].

• Example 3: Rational Functions ➗

  • Given f(x) = (x-5)/(2x+3) and g(x) = (3x+5)/(1-2x). To verify:
    • (f ◦ g)(x) = f(g(x)) = f((3x+5)/(1-2x)) = [((3x+5)/(1-2x)) - 5] / [2((3x+5)/(1-2x)) + 3]
      • Simplify the numerator: (3x+5 - 5(1-2x))/(1-2x) = (3x+5 - 5 + 10x)/(1-2x) = 13x/(1-2x)
      • Simplify the denominator: (2(3x+5) + 3(1-2x))/(1-2x) = (6x+10 + 3 - 6x)/(1-2x) = 13/(1-2x)
      • So, (f ◦ g)(x) = (13x/(1-2x)) / (13/(1-2x)) = x ✅
    • (g ◦ f)(x) = g(f(x)) = g((x-5)/(2x+3)) = [3((x-5)/(2x+3)) + 5] / [1 - 2((x-5)/(2x+3))]
      • Simplify the numerator: (3(x-5) + 5(2x+3))/(2x+3) = (3x-15 + 10x+15)/(2x+3) = 13x/(2x+3)
      • Simplify the denominator: (1(2x+3) - 2(x-5))/(2x+3) = (2x+3 - 2x+10)/(2x+3) = 13/(2x+3)
      • So, (g ◦ f)(x) = (13x/(2x+3)) / (13/(2x+3)) = x ✅
  • These functions are also inverses [5.10].

Logarithmic Functions: A Natural Inverse Example! 🌳↔️📈

The logarithmic function (log_a x) is defined as the inverse of the exponential function (a^x). This is a prime example of an inverse relationship in mathematics!

Practice Questions 📝

Question 1: Let f(x) = 2x - 3 and g(x) = (x + 3) / 2. Are f(x) and g(x) inverse functions? Show your work using the composition rule. 🤔

Question 2: Consider the function h(x) = x². (a) Does h(x) have an inverse function on its entire domain (all real numbers)? Explain why or why not using the Horizontal Line Test. 📏 (b) If not, how could you restrict the domain of h(x) so that it would have an inverse? 🧐

Question 3: If the point (5, 7) is on the graph of a one-to-one function k(x), what point must be on the graph of k⁻¹(x)? 📍

Solutions ✅

Solution 1: To check if f(x) and g(x) are inverse functions, we need to verify if (f ◦ g)(x) = x and (g ◦ f)(x) = x [5.10].

1. Calculate (f ◦ g)(x): (f ◦ g)(x) = f(g(x)) [5.10] = f((x + 3) / 2) = 2 * ((x + 3) / 2) - 3 = (x + 3) - 3 = x ✅

2. Calculate (g ◦ f)(x): (g ◦ f)(x) = g(f(x)) [5.10] = g(2x - 3) = ((2x - 3) + 3) / 2 = (2x) / 2 = x ✅

Since both compositions result in x, yes, f(x) and g(x) are inverse functions [5.10].

Solution 2: (a) h(x) = x². No, h(x) does not have an inverse function on its entire domain (all real numbers) 🚫. Explanation: If we apply the Horizontal Line Test [5.2], a horizontal line (e.g., y = 4) would intersect the graph of h(x) = x² at two points (x = -2 and x = 2). Since different inputs (-2 and 2) produce the same output (4), the function is not one-to-one [1.6.1, 5.2]. Therefore, it cannot have an inverse function on ℝ [5.10].

(b) To make h(x) have an inverse, we need to restrict its domain so that it becomes one-to-one [5.10]. We can do this in a few ways: * Option 1: Restrict the domain to [0, ∞) (all non-negative real numbers). On this domain, h(x) = x² is an increasing function [5.2] and passes the Horizontal Line Test. Its inverse would be h⁻¹(x) = √x. * Option 2: Restrict the domain to (-∞, 0] (all non-positive real numbers). On this domain, h(x) = x² is a decreasing function [5.2] and passes the Horizontal Line Test. Its inverse would be h⁻¹(x) = -√x.

Solution 3: If a point (a, f(a)) is on the graph of a function f, then the point (f(a), a) is on the graph of its inverse function f⁻¹ [5.10]. This means the x and y coordinates are swapped.

Given the point (5, 7) is on the graph of k(x). Here, a = 5 and k(a) = 7. Therefore, the point that must be on the graph of k⁻¹(x) is (7, 5) 📍.