11.2 Functions (Fonksiyonlar): Domain/Range, Composition, Inverses & Functional Equations

Functions (Fonksiyonlar): Types, Composition, Inverses & Functional Equations

Function practice connects algebraic expressions to mappings and graphs. Compare domain and codomain, then use a targeted substitution when asked for one value. Computing a full inverse remains useful when the question asks for its formula or domain.

---

1. Domain, Codomain & Function Types (Fonksiyon Türleri)

A function $f: A \to B$ maps every element of the domain (tanım kümesi) $A$ to exactly one element of the codomain (değer kümesi) $B$. For real-valued functions $f: D \to \mathbb{R}$, the maximal domain $D$ requires:

• Denominators to be nonzero: $\frac{P(x)}{Q(x)} \implies Q(x) \ne 0$.
• Even-index radicals to have non-negative radicands: $\sqrt[2n]{g(x)} \implies g(x) \ge 0$.

If $s(A) = m$ and $s(B) = n$, the total number of functions from $A$ to $B$ is $n^m$, and the number of one-to-one functions (for $m \le n$) is $P(n, m) = \frac{n!}{(n-m)!}$.

Special Function Classifications

1. Injective / One-to-One (Birebir Fonksiyon): $x_1 \ne x_2 \implies f(x_1) \ne f(x_2)$ (every horizontal line intersects the graph at most once).
2. Surjective / Onto (Örten Fonksiyon): $f(A) = B$, meaning no element in $B$ is left unmapped.
3. Bijective (Birebir ve Örten): Both injective and surjective. A function $f: A \to B$ has an inverse function $f^{-1}: B \to A$ if and only if $f$ is bijective.
4. Identity Function (Birim Fonksiyon $I$): Maps every input to itself: $I(x) = x$, and more generally $I(\text{anything}) = \text{anything}$. If $f(x) = (a - 2)x^2 + (b + 1)x + c - 3$ is the identity function, then the coefficient of $x$ is $1$ ($b + 1 = 1 \implies b = 0$) and all other coefficients are $0$ ($a = 2, c = 3$).
5. Constant Function (Sabit Fonksiyon): $f(x) = c$ for all $x$. For a fractional linear expression:

$$f(x) = \frac{ax + b}{cx + d} \text{ is constant} \iff ad-bc=0$$

1. Linear Function (Doğrusal Fonksiyon): $f(x) = ax + b$. If $f$ is stated to be linear and you see a relation like $f(x + 1) + f(x - 2) = 6x + 4$, substitute $f(x) = ax + b$ and equate coefficients.
2. Even & Odd Functions (Çift ve Tek Fonksiyonlar):
   • Even (Çift): $f(-x) = f(x)$ for all $x \in D$. Its graph is symmetric with respect to the $y$-axis ($y$ eksenine göre simetrik). A polynomial is even if and only if all odd-power coefficients are $0$.
   • Odd (Tek): $f(-x) = -f(x)$ for all $x \in D$. Its graph is symmetric with respect to the origin (orijine göre simetrik). A polynomial is odd if and only if all even-power coefficients (including the constant term) are $0$; moreover, if $0 \in D$, then $f(0) = 0$.

---

2. Composition (Bileşke) & Inverse Functions (Ters Fonksiyon)

The composition $(f \circ g)(x) = f(g(x))$ applies $g$ first, then feeds its output into $f$. Composition is associative, $f \circ (g \circ h) = (f \circ g) \circ h$, but generally not commutative ($f \circ g \ne g \circ f$).

The Input-Output Swap Shortcut

Because $f$ and $f^{-1}$ reverse inputs and outputs:

$$f(a) = b \iff f^{-1}(b) = a$$

Linear & Fractional Linear (Möbius) Inverse Shortcuts

When you do need the algebraic formula for $f^{-1}(x)$, memorize these two instant rules:

• Linear Function: For $f(x) = ax + b$ ($a \ne 0$):

$$f^{-1}(x) = \frac{x - b}{a}$$

• Fractional Linear (Möbius) Function: For $f: \mathbb{R} \setminus \{-\frac{d}{c}\} \to \mathbb{R} \setminus \{\frac{a}{c}\}$ defined by:

$$f(x) = \frac{ax + b}{cx + d} \implies f^{-1}(x) = \frac{-dx + b}{cx - a}$$

Simply swap and negate the top-left coefficient $a$ and the bottom-right constant $d$, while leaving $b$ and $c$ untouched! Notice also why the excluded domain value $-\frac{d}{c}$ makes the denominator of $f$ zero, while the excluded codomain value $\frac{a}{c}$ makes the denominator of $f^{-1}$ zero.

Inverse of a Composition (Bileşkenin Tersi)

Taking the inverse of a composition reverses the order of the functions (just like taking off shoes and socks):

$$(f \circ g)^{-1}(x) = (g^{-1} \circ f^{-1})(x) = g^{-1}(f^{-1}(x))$$

Because $f \circ f^{-1} = f^{-1} \circ f = I$, you can isolate an unknown function inside a composition by composing both sides on the appropriate side with an inverse. For example, if $(f \circ g)(x) = h(x)$, then:

• $f(x) = (h \circ g^{-1})(x) = h(g^{-1}(x))$ (compose on the right with $g^{-1}$)
• $g(x) = (f^{-1} \circ h)(x) = f^{-1}(h(x))$ (compose on the left with $f^{-1}$)

---

3. Telescoping Functional Equations (İndirgemeli Fonksiyonlar)

When a functional equation relates $f(x + 1)$ (or $f(x + k)$) to $f(x)$ and asks for $f(b)$ given $f(a)$, write out the equation step-by-step from $x = a$ to $x = b - 1$ and telescope:

1. Additive Recurrence ($f(x + 1) - f(x) = g(x)$): Write the equations vertically for $x = a, a + 1, \dots, b - 1$ and add column-wise. All intermediate terms $+f(k)$ and $-f(k)$ cancel, leaving:

$$f(b) - f(a) = \sum_{x=a}^{b-1} g(x)$$

1. Multiplicative Recurrence ($\frac{f(x + 1)}{f(x)} = h(x)$ or $f(x + 1) = h(x) \cdot f(x)$): Write the equations for $x = a, a + 1, \dots, b - 1$ and multiply column-wise. All intermediate factors cancel on opposite sides, leaving:

$$\frac{f(b)}{f(a)} = \prod_{x=a}^{b-1} h(x)$$

---

Summary Reference Table

---

Worked Examples

Worked Example 1: Constant Rational Condition & Möbius Inverse

Problem: The function $g(x) = \frac{(m - 2)x + 6}{2x - 3}$ is a constant function on its domain. Using this value of $m$, let $h(x) = \frac{mx + 1}{2x - 5}$. Find $h^{-1}(3)$.

Solution: Since $g(x)$ is constant, the ratio of the $x$-coefficients equals the ratio of the constant terms:

$$\frac{m - 2}{2} = \frac{6}{-3} = -2 \implies m - 2 = -4 \implies m = -2$$

Substituting $m = -2$ into $h(x)$ gives $h(x) = \frac{-2x + 1}{2x - 5}$. Applying the Möbius inverse rule (swap and negate $a = -2$ and $d = -5$), we get:

$$h^{-1}(x) = \frac{5x + 1}{2x + 2} \implies h^{-1}(3) = \frac{5(3) + 1}{2(3) + 2} = \frac{16}{8} = \mathbf{2}$$

(Alternatively, set $\frac{-2x + 1}{2x - 5} = 3 \implies -2x + 1 = 6x - 15 \implies 8x = 16 \implies x = 2$.)

Worked Example 2: Additive & Multiplicative Telescoping Recurrences

Problem:

1. Given $f(x + 1) - f(x) = 2x + 3$ for all integers $x$ and $f(1) = 4$, find $f(15)$.
2. Given $g(x + 1) = \frac{x + 2}{x} \cdot g(x)$ for integers $x \ge 1$ and $g(1) = 3$, find $g(10)$.

Solution:

1. Summing $f(x + 1) - f(x) = 2x + 3$ from $x = 1$ to $x = 14$ telescopes the left side to $f(15) - f(1)$:

$$f(15) - 4 = \sum_{x=1}^{14} (2x + 3) = 2\left(\frac{14 \cdot 15}{2}\right) + 3(14) = 210 + 42 = 252 \implies f(15) = \mathbf{256}$$

1. Multiplying $\frac{g(x+1)}{g(x)} = \frac{x+2}{x}$ for $x = 1, 2, \dots, 9$ gives:

$$\frac{g(10)}{g(1)} = \frac{3}{1} \cdot \frac{4}{2} \cdot \frac{5}{3} \cdot \frac{6}{4} \cdots \frac{11}{9} = \frac{10 \cdot 11}{1 \cdot 2} = 55 \implies g(10) = 3 \cdot 55 = \mathbf{165}$$

The Möbius inverse and its displayed excluded values require $c\ne0$ and $ad-bc\ne0$. If $c=0$, handle the linear or constant case separately. Even/odd function definitions require a domain symmetric about zero. Multiplicative telescoping by ratios requires nonzero terms; the product recurrence $f(b)=f(a)\prod h(x)$ avoids dividing by $f(a)$.

Review the Method

Telescoping functional equations of the form f(x + 1) - f(x) = g(x) are solved by column-wise addition across consecutive inputs, while f(x + 1) = h(x) · f(x) is solved by column-wise multiplication.