Opérations sur les Fonctions et Résolution d'Exercices Complexes

TL;DR

You'll learn how to combine functions using basic operations like addition, subtraction, multiplication, and division. We'll also cover function composition, which involves applying one function's output as another's input. Finally, we'll tackle inverse functions and how to solve problems involving these concepts.

1. The Mental Model

Think of functions as mini-machines: you put something in, and they give you something out. Operations on functions are like linking these machines together or creating new ones from existing parts.

2. The Core Material

When you operate on functions, you're essentially performing arithmetic or composition with their output values.

Addition, Subtraction, Multiplication, and Division of Functions

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Let $f(x)$ and $g(x)$ be two functions.

• Addition: $(f+g)(x) = f(x) + g(x)$
  • Example: If $f(x) = 2x$ and $g(x) = x+1$, then $(f+g)(x) = 2x + (x+1) = 3x+1$.
• Subtraction: $(f-g)(x) = f(x) - g(x)$
  • Example: If $f(x) = 2x$ and $g(x) = x+1$, then $(f-g)(x) = 2x - (x+1) = x-1$.
• Multiplication: $(f \cdot g)(x) = f(x) \cdot g(x)$
  • Example: If $f(x) = 2x$ and $g(x) = x+1$, then $(f \cdot g)(x) = 2x(x+1) = 2x^2 + 2x$.
• Division: $(f/g)(x) = f(x) / g(x)$, where $g(x) \neq 0$
  • Example: If $f(x) = 2x$ and $g(x) = x+1$, then $(f/g)(x) = \frac{2x}{x+1}$, for $x \neq -1$.

Domain Considerations: The domain of these combined functions is the intersection of the domains of $f$ and $g$, with the added condition for division that the denominator isn't zero.

Function Composition

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Function composition means applying one function to the result of another. It's like chaining machines together.

• Notation: $(f \circ g)(x)$ or $f(g(x))$
• Meaning: First, you calculate $g(x)$, and then you use that result as the input for $f$.
• Order matters! $(f \circ g)(x)$ is generally not the same as $(g \circ f)(x)$.

graph TD
    A["Input x"] --> B["Function g(x)"]
    B --> C["Output g(x)"]
    C --> D["Function f(y) where y = g(x)"]
    D --> E["Output f(g(x))"]

• Example: If $f(x) = x^2$ and $g(x) = x+3$:
  • $(f \circ g)(x) = f(g(x)) = f(x+3) = (x+3)^2 = x^2 + 6x + 9$
  • $(g \circ f)(x) = g(f(x)) = g(x^2) = x^2 + 3$
    Notice how the results are different.

Inverse Functions

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An inverse function "undoes" what the original function does. If $f(a) = b$, then its inverse, denoted $f^{-1}(b) = a$.

• Key Property: $(f \circ f^{-1})(x) = x$ and $(f^{-1} \circ f)(x) = x$.

• How to find an inverse:

  1. Replace $f(x)$ with $y$.
  2. Swap $x$ and $y$.
  3. Solve the new equation for $y$.
  4. Replace $y$ with $f^{-1}(x)$.

• Example: Find the inverse of $f(x) = 2x+1$.

  1. $y = 2x+1$
  2. $x = 2y+1$
  3. $x-1 = 2y \implies y = \frac{x-1}{2}$
  4. $f^{-1}(x) = \frac{x-1}{2}$

Important: A function must be one-to-one (each output comes from only one input) to have an inverse. Graphically, this means it passes the horizontal line test.

3. Worked Example

Let's work through an example combining composition and inverse functions.

Given $f(x) = 3x - 2$ and $g(x) = \sqrt{x+1}$.
Find $(f \circ g)(x)$ and the inverse of $f(x)$, denoted $f^{-1}(x)$.

Part 1: Find $(f \circ g)(x)$

1. Remember $(f \circ g)(x) = f(g(x))$.
2. Substitute $g(x)$ into $f(x)$:
   $f(g(x)) = f(\sqrt{x+1})$
3. Now replace $x$ in $f(x)$ with $\sqrt{x+1}$:
   $f(\sqrt{x+1}) = 3(\sqrt{x+1}) - 2$

So, $(f \circ g)(x) = 3\sqrt{x+1} - 2$.
Domain for $(f \circ g)(x)$: Since $g(x)$ involves a square root, $x+1 \ge 0 \implies x \ge -1$. The domain of $f$ is all real numbers, so the domain of $(f \circ g)(x)$ is $x \ge -1$.

Part 2: Find $f^{-1}(x)$

1. Start with $f(x) = 3x - 2$. Replace $f(x)$ with $y$:
   $y = 3x - 2$
2. Swap $x$ and $y$:
   $x = 3y - 2$
3. Solve for $y$:
   $x + 2 = 3y$
   $y = \frac{x+2}{3}$
4. Replace $y$ with $f^{-1}(x)$:
   $f^{-1}(x) = \frac{x+2}{3}$

4. Key Takeaways

• Function operations (addition, subtraction, multiplication, division) combine outputs of functions at the same input.
• Function composition $f(g(x))$ means applying $g$ first, then $f$ to the result. Order matters!
• The domain of combined or composed functions must respect the restrictions of all involved functions.
• An inverse function $f^{-1}(x)$ "undoes" $f(x)$, meaning $f(f^{-1}(x)) = x$.
• To find an inverse, replace $f(x)$ with $y$, swap $x$ and $y$, then solve for $y$.
• Only one-to-one functions have a true inverse.
• Be careful with domain restrictions, especially for square roots and division by zero.

5. Now Try It

For $f(x) = x^2 - 4$ and $g(x) = x-2$:
1. Calculate $(f/g)(x)$ and specify its domain.
2. Calculate $(g \circ f)(x)$.
3. Find $f^{-1}(x)$ for $x \ge 0$.

What to do: Follow the steps for each operation. For the inverse, remember you might need to restrict the domain of the original function to make it one-to-one.
What success looks like: You should get $(f/g)(x) = x+2$ with domain $x \neq 2$, $(g \circ f)(x) = x^2 - 6$, and $f^{-1}(x) = \sqrt{x+4}$ for $x \ge -4$.