Understanding Bijective Functions and Function Composition

Exam Tip: When proving surjectivity for rational functions, always check if the $x$ you solved for is actually defined within the domain. For example, if your domain excludes a certain value, you must prove that your formula for $x$ never produces that excluded value.

Worked Example: Proving Bijectivity

Consider the function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = 4x - 3$.

1. Prove Injectivity: Assume $f(x_1) = f(x_2)$:

$$\begin{aligned} 4x_1 - 3 &= 4x_2 - 3 \\ 4x_1 &= 4x_2 \\ x_1 &= x_2 \end{aligned}$$

Since $x_1 = x_2$, the function is injective.

2. Prove Surjectivity: Let $y \in \mathbb{R}$. We set $y = 4x - 3$ and solve for $x$:

$$\begin{aligned} y + 3 &= 4x \\ x &= \frac{y + 3}{4} \end{aligned}$$

Since $y$ is a real number, $\frac{y+3}{4}$ is also a real number. Plugging this back into the function:

$$\begin{aligned} f\left(\frac{y+3}{4}\right) &= 4\left(\frac{y+3}{4}\right) - 3 \\ &= (y + 3) - 3 \\ &= y \end{aligned}$$

Thus, $f$ is surjective. Since it is both injective and surjective, $f$ is bijective.

The Inverse Function

Once a function is proven to be bijective, we know it is invertible. This means there exists an inverse function, denoted $f^{-1}$, that "undoes" the action of $f$. If $f$ maps $A$ to $B$, then $f^{-1}$ maps $B$ back to $A$.

To find the rule for $f^{-1}(x)$, you solve the equation $y = f(x)$ for $x$ in terms of $y$, and then replace $y$ with $x$.

Example: For the complex mapping $f(x) = \frac{4x-5}{3x-7}$ where $f: \mathbb{R}-\{7/3\} \to \mathbb{R}-\{4/3\}$, the inverse is found by rearranging the terms to get: $f^{-1}(x) = \frac{7x-5}{3x-4}$.

Function Composition

Function composition is the process of applying one function to the result of another. If we have two functions, $f$ and $g$, the composition $(f \circ g)(x)$ is defined as $f(g(x))$.

Key Rules of Composition

1. Order Matters: Composition is generally not commutative. This means $(f \circ g)(x)$ is usually not the same as $(g \circ f)(x)$.
2. Inside-Out Evaluation: When dealing with multiple functions, such as $(g \circ h \circ f)(x)$, you evaluate from the innermost function outward: first $f(x)$, then apply $h$ to that result, and finally apply $g$.
3. Identity Relationship: A special case occurs when $f$ and $g$ are inverses. In this case, composing them results in the identity function: $(f \circ g)(x) = x$.

Worked Example: Multiple Compositions

Let $f(x) = x + 3$, $g(x) = x - 3$, and $h(x) = 3x$.

Find $(g \circ h)(x)$:

$$\begin{aligned} (g \circ h)(x) &= g(h(x)) \\ &= g(3x) \\ &= 3x - 3 \end{aligned}$$

Find $(g \circ h \circ f)(x)$: First, find $h(f(x))$:

$$\begin{aligned} h(f(x)) &= h(x + 3) \\ &= 3(x + 3) \\ &= 3x + 9 \end{aligned}$$

Then, apply $g$ to that result:

$$\begin{aligned} g(3x + 9) &= (3x + 9) - 3 \\ &= 3x + 6 \end{aligned}$$

Thus, $(g \circ h \circ f)(x) = 3x + 6$.