Inverse Functions and Graph Transformations (Video Notes)
Inverse Functions: Core Concepts
Inverse functions concept: If a function f has a point (a, b) on its graph, then its inverse function f^{-1} will have the point (b, a) on its graph. This fundamentally means that inverse functions swap inputs and outputs: the output of the original function becomes the input of the inverse, and the input of the original becomes the output of the inverse.
Example: Consider the function f(x) = 2x + 3. If we input x=1, the output is f(1) = 2(1) + 3 = 5. So, the point (1, 5) is on the graph of f(x). For its inverse, f^{-1}(x), the point (5, 1) must be on its graph. This implies that f^{-1}(5) = 1.
Algebraically, if y = f(x), then the inverse satisfies x = f^{-1}(y). This can also be expressed through the composition of functions: f^{-1}(f(x)) = x for all x in the domain of f, and f(f^{-1}(x)) = x for all x in the domain of f^{-1}.
Finding an Inverse Example: To find the inverse of f(x) = 2x + 3:
Replace f(x) with y: y = 2x + 3
Swap x and y: x = 2y + 3
Solve for y:
x - 3 = 2y
y = \frac{x - 3}{2}Replace y with f^{-1}(x): f^{-1}(x) = \frac{x - 3}{2}
Graphically, the graph of f^{-1} is the reflection of the graph of f across the line y = x. Every point (a, b) on f becomes (b, a) on f^{-1}, and the line y=x acts as the mirror between them.
Key property for checking inverse pairs
To check if two functions f and g are inverses, verify that the compositions yield the identity function:
(f \circ g)(x) = f(g(x)) = x \text{ and } (g \circ f)(x) = g(f(x)) = xIf both of these conditions hold (on the appropriate domains), then g = f^{-1} and vice versa.
Composition Check Example: Using f(x) = 2x + 3 and f^{-1}(x) = \frac{x - 3}{2}.
(f \circ f^{-1})(x) = f(f^{-1}(x)) = f\left(\frac{x - 3}{2}\right) = 2\left(\frac{x - 3}{2}\right) + 3 = (x - 3) + 3 = x
(f^{-1} \circ f)(x) = f^{-1}(f(x)) = f^{-1}(2x + 3) = \frac{(2x + 3) - 3}{2} = \frac{2x}{2} = x
Since both compositions result in x, f(x) and f^{-1}(x) are indeed inverses.
Example: solving 1/x^2 = 5 and implications for inverses
Start with \frac{1}{x^2} = 5
Multiply both sides by x^2 (assuming x \ne 0): 1 = 5x^2
Divide by 5: x^2 = \frac{1}{5}
Thus x = \pm\frac{1}{\sqrt{5}}
Important caveat: The function h(x) = \frac{1}{x^2} is not one-to-one on its entire domain (all real numbers except x = 0). For instance, h(2) = \frac{1}{4} and h(-2) = \frac{1}{4}. Since different inputs (2 and -2) lead to the same output (1/4), the function is not one-to-one, and therefore does not have a single inverse unless we restrict the domain.
Domain Restriction Example: If we restrict the domain of h(x) = \frac{1}{x^2} to x > 0, then it becomes one-to-one. For this restricted domain, we can find an inverse:
y = \frac{1}{x^2}
Swap x and y: x = \frac{1}{y^2}
Solve for y: y^2 = \frac{1}{x} \implies y = \sqrt{\frac{1}{x}} = \frac{1}{\sqrt{x}}
So, for x > 0, the inverse is h^{-1}(x) = \frac{1}{\sqrt{x}}, where the domain of h^{-1}(x) would be x > 0. Note that we take only the positive square root to align with the restricted domain of the original function.
Connections to foundational principles
One-to-one functions and domain restrictions: An inverse function can only exist as a proper function if the original function is one-to-one. A function is one-to-one if every output corresponds to exactly one input. Graphically, this is verified by the Horizontal Line Test: if any horizontal line intersects the function's graph at more than one point, the function is not one-to-one. If a function is not naturally one-to-one, we must restrict its domain (e.g., x \ge 0 for y = x^2) to make it one-to-one and allow an inverse to be defined for that specific portion.
Symmetry and reflections: As mentioned, the relationship between a function and its inverse is one of symmetry across the line y = x. If you imagine folding the coordinate plane along the line y=x, the graph of f would perfectly superimpose on the graph of f^{-1}. For instance, take f(x) = x+1. The point (0,1) is on its graph. The inverse is f^{-1}(x) = x-1, and the point (1,0) is on its graph. These are reflections across y=x.
Inverse operations as a core problem-solving tool: The concept of inverse functions is deeply tied to the idea of undoing an operation to solve for an unknown. When you solve an equation like 2x+3=7, you are essentially applying inverse operations (subtracting 3, then dividing by 2) to isolate x. These operations correspond to the steps taken to find the inverse function. Many real-world problems involve finding the input that produced a given output, which is precisely what an inverse function allows us to do.
Linear models as a first bridge to understanding rate of change: Linear models, in the form y = mx + b, provide a fundamental example for understanding inverses. For example, converting temperature from Celsius (C) to Fahrenheit (F) is given by F = \frac{9}{5}C + 32. To find the inverse (converting Fahrenheit to Celsius), we solve for C:
F - 32 = \frac{9}{5}C
C = \frac{5}{9}(F - 32)
This inverse relationship is itself a linear model, illustrating how inverse functions 'undo' the original process.
Practical numerical handling: Pushing through equations by isolating the desired variable with inverse operations is a central skill in algebra and directly translates to understanding inverse function theory. The process of "doing and undoing" operations is the essence of both solving equations and finding inverse functions, underscoring their practical importance in mathematical workflows.