1.2 MCQs-Functions

Functions

Basic Concepts and Definitions

1. A function f from set A to set B is defined as:

1. A relation where each element of A is related to at least one element of B

2. A relation where each element of A is related to exactly one element of B

3. A relation where each element of B is related to at least one element of A

4. A relation where each element of B is related to exactly one element of A

Show me the answer

Answer: 2. A relation where each element of A is related to exactly one element of B

Explanation:

• A function $f: A \to B$ is a special type of relation where:

  • For every element $a \in A$, there exists exactly one element $b \in B$ such that $(a, b) \in f$.

• This is often written as $f(a) = b$.

• The set A is called the domain, and B is called the codomain.

• The range is the set of all actual output values: $\{f(a) : a \in A\} \subseteq B$.

2. Which of the following relations from $A = \{1, 2, 3\}$ to $B = \{4, 5, 6\}$ represents a function?

1. $\{(1, 4), (2, 5)\}$

2. $\{(1, 4), (2, 5), (3, 6), (1, 5)\}$

3. $\{(1, 4), (2, 5), (3, 6)\}$

4. $\{(1, 4), (2, 4), (3, 4)\}$

Show me the answer

Answer: 3. $\{(1, 4), (2, 5), (3, 6)\}$

Explanation:

• For a relation to be a function from A to B:

  1. Every element of A must appear as the first element in at least one ordered pair.

  2. No element of A can appear as the first element in more than one ordered pair.

• Option 1 fails condition 1: element 3 from A is missing.

• Option 2 fails condition 2: element 1 from A appears twice (with both 4 and 5).

• Option 3 satisfies both conditions: each of 1,2,3 appears exactly once as first element.

• Option 4 also satisfies both conditions (it's a constant function).

3. The domain of a function is:

1. The set of all possible input values

2. The set of all possible output values

3. The set of all ordered pairs (x, f(x))

4. The set of all points where the function is defined

Show me the answer

Answer: 1. The set of all possible input values

Explanation:

• For a function $f: A \to B$:

  • Domain = A = set of all input values.

  • Codomain = B = set of all possible output values.

  • Range = $\{f(x) : x \in A\}$ = actual set of output values.

• The domain consists of all values for which the function is defined and produces an output.

• In many practical problems, the domain is determined by restrictions like:

  • Denominator cannot be zero

  • Square root argument must be non-negative

  • Logarithm argument must be positive

4. The range of the function $f(x) = x^2$ with domain $\{ -2, -1, 0, 1, 2 \}$ is:

1. $\{ -4, -1, 0, 1, 4 \}$

2. $\{ 0, 1, 4 \}$

3. $\{ 0, 1, 2, 4 \}$

4. $\{ -2, -1, 0, 1, 2 \}$

Show me the answer

Answer: 2. $\{ 0, 1, 4 \}$

Explanation:

• Range is the set of actual output values obtained by applying the function to all elements in the domain.

• Given $f(x) = x^2$ and domain = $\{ -2, -1, 0, 1, 2 \}$

• Calculate: $f(-2) = (-2)^2 = 4$ $f(-1) = (-1)^2 = 1$ $f(0) = 0^2 = 0$ $f(1) = 1^2 = 1$ $f(2) = 2^2 = 4$

• Distinct output values: 0, 1, 4

• Therefore, range = $\{ 0, 1, 4 \}$

• Note: The range is a set, so duplicates are not listed.

Types of Functions

5. A function $f: A \to B$ is said to be one-to-one (injective) if:

1. Every element of B has at least one pre-image in A

2. Different elements of A have different images in B

3. Every element of B has exactly one pre-image in A

4. The range equals the codomain

Show me the answer

Answer: 2. Different elements of A have different images in B

Explanation:

• A function is injective (one-to-one) if it preserves distinctness: $\text{If } a_1 \neq a_2 \text{, then } f(a_1) \neq f(a_2)$

• Equivalently: $\text{If } f(a_1) = f(a_2) \text{, then } a_1 = a_2$

• Injective functions never map two different domain elements to the same codomain element.

• Examples:

  • $f(x) = 2x$ is injective (different x give different 2x)

  • $f(x) = x^2$ from $\mathbb{R}$ to $\mathbb{R}$ is NOT injective (both 2 and -2 map to 4)

6. A function $f: A \to B$ is said to be onto (surjective) if:

1. Every element of B has at least one pre-image in A

2. Different elements of A have different images in B

3. Every element of B has exactly one pre-image in A

4. The domain equals the codomain

Show me the answer

Answer: 1. Every element of B has at least one pre-image in A

Explanation:

• A function is surjective (onto) if its range equals its codomain.

• Formally: For every $b \in B$, there exists at least one $a \in A$ such that $f(a) = b$.

• Surjective functions "cover" the entire codomain.

• Examples:

  • $f(x) = 2x$ from $\mathbb{R}$ to $\mathbb{R}$ is surjective

  • $f(x) = x^2$ from $\mathbb{R}$ to $[0, \infty)$ is surjective

  • $f(x) = x^2$ from $\mathbb{R}$ to $\mathbb{R}$ is NOT surjective (negative numbers have no pre-image)

7. A function that is both one-to-one and onto is called:

1. Injective

2. Surjective

3. Bijective

4. Constant

Show me the answer

Answer: 3. Bijective

Explanation:

• A bijective function is both injective (one-to-one) and surjective (onto).

• Properties of bijective functions:

  1. Every element of the domain maps to a unique element of the codomain.

  2. Every element of the codomain has exactly one pre-image.

• Bijective functions establish a one-to-one correspondence between domain and codomain.

• If $f: A \to B$ is bijective, then:

  • $|A| = |B|$ (sets have the same cardinality)

  • The inverse function $f^{-1}: B \to A$ exists.

8. Which of the following functions is injective but not surjective?

1. $f: \mathbb{R} \to \mathbb{R}, f(x) = x^3$

2. $f: \mathbb{R} \to \mathbb{R}, f(x) = x^2$

3. $f: \mathbb{R} \to [0, \infty), f(x) = x^2$

4. $f: \mathbb{R} \to \mathbb{R}, f(x) = e^x$

Show me the answer

Answer: 4. $f: \mathbb{R} \to \mathbb{R}, f(x) = e^x$

Explanation:

• Let's analyze each option:

  1. $f(x) = x^3$ is both injective and surjective on $\mathbb{R}$, so it's bijective.

  2. $f(x) = x^2$ is neither injective (e.g., f(2)=f(-2)=4) nor surjective (negative numbers not in range).

  3. $f(x) = x^2$ with codomain $[0, \infty)$ is surjective but not injective.

  4. $f(x) = e^x$:

     • Injective: If $e^{x_1} = e^{x_2}$, then $x_1 = x_2$.

     • Not surjective: Range is $(0, \infty)$, which is not all of $\mathbb{R}$ (negative numbers and zero have no pre-image).

9. A function where all elements of the domain map to the same element in the codomain is called:

1. Identity function

2. Constant function

3. Bijective function

4. Onto function

Show me the answer

Answer: 2. Constant function

Explanation:

• A constant function has the same output value for every input.

• Form: $f(x) = c$ for all $x \in A$, where c is a fixed element of B.

• Examples:

  • $f: \mathbb{R} \to \mathbb{R}, f(x) = 5$ for all x

  • $f: \{1,2,3\} \to \{a,b\}, f(1)=f(2)=f(3)=a$

• Properties:

  • Never injective (unless domain has only one element)

  • Surjective only if codomain = $\{c\}$ (singleton set)

  • Graph is a horizontal line (for real functions)

10. The identity function on a set A is defined as:

1. $f(x) = 0$ for all $x \in A$

2. $f(x) = 1$ for all $x \in A$

3. $f(x) = x$ for all $x \in A$

4. $f(x) = -x$ for all $x \in A$

Show me the answer

Answer: 3. $f(x) = x$ for all $x \in A$

Explanation:

• The identity function $I_A: A \to A$ maps each element to itself.

• Notation: $I_A(x) = x$ for all $x \in A$.

• Properties:

  • Always bijective (both injective and surjective)

  • Its inverse is itself: $I_A^{-1} = I_A$

  • Composition with any function f: $f \circ I_A = I_A \circ f = f$

• The identity function is the neutral element for function composition.

Composition of Functions

11. If $f: A \to B$ and $g: B \to C$ are functions, then the composition $g \circ f$ is:

1. A function from A to C defined by $(g \circ f)(x) = g(f(x))$

2. A function from B to A defined by $(g \circ f)(x) = f(g(x))$

3. A function from A to C defined by $(g \circ f)(x) = f(g(x))$

4. A function from C to A defined by $(g \circ f)(x) = g(f(x))$

Show me the answer

Answer: 1. A function from A to C defined by $(g \circ f)(x) = g(f(x))$

Explanation:

• Function composition $g \circ f$ is defined only when the codomain of f equals the domain of g.

• The composition is read "g circle f" or "g of f".

• Order matters: $g \circ f$ means apply f first, then g.

• Domain of $g \circ f$ = domain of f = A.

• Codomain of $g \circ f$ = codomain of g = C.

• Example: If $f(x) = x^2$ and $g(x) = x + 1$, then: $(g \circ f)(x) = g(f(x)) = g(x^2) = x^2 + 1$

12. For functions $f: A \to B$ and $g: B \to C$, if both f and g are injective, then $g \circ f$ is:

1. Injective

2. Surjective

3. Bijective

4. Not necessarily injective

Show me the answer

Answer: 1. Injective

Explanation:

• If f and g are both injective, then their composition $g \circ f$ is also injective.

• Proof: Suppose $(g \circ f)(x_1) = (g \circ f)(x_2)$. Then $g(f(x_1)) = g(f(x_2))$. Since g is injective, $f(x_1) = f(x_2)$. Since f is injective, $x_1 = x_2$. Therefore, $g \circ f$ is injective.

• However, $g \circ f$ may not be surjective even if both f and g are surjective.

• Example: f: ℝ→(0,∞) with f(x)=e^x (injective), g: (0,∞)→ℝ with g(x)=ln(x) (injective). Then g∘f(x)=ln(e^x)=x is the identity on ℝ, which is injective.

13. For functions $f: A \to B$ and $g: B \to C$, if both f and g are surjective, then $g \circ f$ is:

1. Injective

2. Surjective

3. Bijective

4. Not necessarily surjective

Show me the answer

Answer: 2. Surjective

Explanation:

• If f and g are both surjective, then their composition $g \circ f$ is also surjective.

• Proof: Let $c \in C$ be arbitrary. Since g is surjective, there exists $b \in B$ such that $g(b) = c$. Since f is surjective, there exists $a \in A$ such that $f(a) = b$. Then $(g \circ f)(a) = g(f(a)) = g(b) = c$. Therefore, every $c \in C$ has a pre-image in A, so $g \circ f$ is surjective.

• However, $g \circ f$ may not be injective even if both f and g are injective.

• Example: f: ℝ→(0,∞) with f(x)=e^x (surjective onto (0,∞)), g: (0,∞)→ℝ with g(x)=ln(x) (surjective onto ℝ). Then g∘f(x)=ln(e^x)=x is the identity on ℝ, which is surjective.

14. For any function $f: A \to A$, $f \circ I_A$ equals:

1. $I_A$

2. $f$

3. $f^{-1}$

4. $I_A \circ f$

Show me the answer

Answer: 2. $f$

Explanation:

• The identity function $I_A$ acts as the identity element for function composition.

• For any function $f: A \to A$: $f \circ I_A = f$ and $I_A \circ f = f$

• Proof: For any $x \in A$, $(f \circ I_A)(x) = f(I_A(x)) = f(x)$ $(I_A \circ f)(x) = I_A(f(x)) = f(x)$

• This is analogous to multiplying by 1 in arithmetic: $a \times 1 = 1 \times a = a$.

Inverse Functions

15. A function $f: A \to B$ has an inverse function $f^{-1}: B \to A$ if and only if:

1. f is injective

2. f is surjective

3. f is bijective

4. f is constant

Show me the answer

Answer: 3. f is bijective

Explanation:

• A function has an inverse function if and only if it is bijective (both injective and surjective).

• Requirements for an inverse:

  1. Injective: Ensures $f^{-1}$ is a function (each input maps to a unique output).

  2. Surjective: Ensures domain of $f^{-1}$ is all of B.

• Properties of inverse functions: $f^{-1}(f(x)) = x$ for all $x \in A$ $f(f^{-1}(y)) = y$ for all $y \in B$ $(f^{-1})^{-1} = f$

• Example: $f: \mathbb{R} \to \mathbb{R}, f(x) = 2x + 3$ is bijective, so it has inverse $f^{-1}(x) = \frac{x-3}{2}$.

16. If $f(x) = 2x + 3$, then its inverse $f^{-1}(x)$ is:

1. $\frac{x-3}{2}$

2. $\frac{x+3}{2}$

3. $2x - 3$

4. $\frac{1}{2x+3}$

Show me the answer

Answer: 1. $\frac{x-3}{2}$

Explanation:

• To find the inverse of $f(x) = 2x + 3$:

  1. Write $y = 2x + 3$

  2. Solve for x in terms of y: $2x = y - 3$, so $x = \frac{y-3}{2}$

  3. Swap x and y: $y = \frac{x-3}{2}$

• Therefore, $f^{-1}(x) = \frac{x-3}{2}$

• Verification: $f^{-1}(f(x)) = f^{-1}(2x+3) = \frac{(2x+3)-3}{2} = \frac{2x}{2} = x$ $f(f^{-1}(x)) = f\left(\frac{x-3}{2}\right) = 2\left(\frac{x-3}{2}\right) + 3 = (x-3) + 3 = x$

17. For a bijective function f, $(f^{-1} \circ f)(x)$ equals:

1. $f(x)$

2. $f^{-1}(x)$

3. x

4. 1

Show me the answer

Answer: 3. x

Explanation:

• For a bijective function $f: A \to B$ with inverse $f^{-1}: B \to A$: $(f^{-1} \circ f)(x) = f^{-1}(f(x)) = x$ for all $x \in A$ $(f \circ f^{-1})(y) = f(f^{-1}(y)) = y$ for all $y \in B$

• This is the defining property of inverse functions.

• Geometrically, the graph of $f^{-1}$ is the reflection of the graph of f about the line $y = x$.

• Example: For $f(x) = 2x$, $f^{-1}(x) = \frac{x}{2}$: $f^{-1}(f(x)) = f^{-1}(2x) = \frac{2x}{2} = x$

18. The inverse of the identity function $I_A: A \to A$ is:

1. The constant function

2. The zero function

3. $I_A$ itself

4. Does not exist

Show me the answer

Answer: 3. $I_A$ itself

Explanation:

• The identity function $I_A(x) = x$ is its own inverse.

• Since $I_A$ is bijective, it has an inverse.

• Verification: $I_A^{-1}(I_A(x)) = I_A^{-1}(x) = x$ $I_A(I_A^{-1}(x)) = I_A(x) = x$

• This shows that $I_A^{-1} = I_A$.

• In general, a function that is its own inverse is called an involution.

Special Functions

19. The greatest integer function (floor function) $f(x) = \lfloor x \rfloor$ gives:

1. The smallest integer ≥ x

2. The largest integer ≤ x

3. The integer closest to x

4. The absolute value of x

Show me the answer

Answer: 2. The largest integer ≤ x

Explanation:

• The greatest integer function or floor function $\lfloor x \rfloor$ gives the greatest integer less than or equal to x.

• Also called the "integer part" of x.

• Examples: $\lfloor 3.7 \rfloor = 3$ $\lfloor -2.3 \rfloor = -3$ (not -2, because -3 ≤ -2.3) $\lfloor 5 \rfloor = 5$

• The ceiling function $\lceil x \rceil$ gives the smallest integer ≥ x.

• Properties:

  • $\lfloor x \rfloor ≤ x < \lfloor x \rfloor + 1$

  • Graph consists of horizontal steps.

20. The absolute value function $f(x) = |x|$ is defined as:

1. $|x| = x$ for all x

2. $|x| = -x$ for all x

3. $|x| = \begin{cases} x & \text{if } x ≥ 0 \\ -x & \text{if } x < 0 \end{cases}$

4. $|x| = \sqrt{x^2}$

Show me the answer

Answer: 3. $|x| = \begin{cases} x & \text{if } x ≥ 0 \\ -x & \text{if } x < 0 \end{cases}$

Explanation:

• The absolute value of a real number x, denoted |x|, is its distance from 0 on the number line.

• Equivalent definitions:

  1. Piecewise: $|x| = \begin{cases} x & \text{if } x ≥ 0 \\ -x & \text{if } x < 0 \end{cases}$

  2. Using square root: $|x| = \sqrt{x^2}$

• Properties:

  • $|x| ≥ 0$ for all x

  • $|x| = 0$ if and only if x = 0

  • $|xy| = |x| \cdot |y|$

  • $|x+y| ≤ |x| + |y|$ (triangle inequality)

• Graph: V-shaped, symmetric about the y-axis.

21. A function f is called even if:

1. $f(-x) = f(x)$ for all x in the domain

2. $f(-x) = -f(x)$ for all x in the domain

3. $f(x+y) = f(x) + f(y)$ for all x, y

4. $f(xy) = f(x)f(y)$ for all x, y

Show me the answer

Answer: 1. $f(-x) = f(x)$ for all x in the domain

Explanation:

• A function f is even if it satisfies $f(-x) = f(x)$ for all x in its domain.

• Geometric interpretation: The graph is symmetric with respect to the y-axis.

• Examples:

  • $f(x) = x^2$ (check: $f(-x) = (-x)^2 = x^2 = f(x)$)

  • $f(x) = \cos x$

  • $f(x) = |x|$

• Properties:

  • If f and g are even, then f+g, f-g, and f·g are even.

  • The composition of two even functions is even.

  • The only function that is both even and odd is the zero function.

22. A function f is called odd if:

1. $f(-x) = f(x)$ for all x in the domain

2. $f(-x) = -f(x)$ for all x in the domain

3. $f(x+y) = f(x) + f(y)$ for all x, y

4. $f(xy) = f(x)f(y)$ for all x, y

Show me the answer

Answer: 2. $f(-x) = -f(x)$ for all x in the domain

Explanation:

• A function f is odd if it satisfies $f(-x) = -f(x)$ for all x in its domain.

• Geometric interpretation: The graph is symmetric with respect to the origin (180° rotation).

• Examples:

  • $f(x) = x^3$ (check: $f(-x) = (-x)^3 = -x^3 = -f(x)$)

  • $f(x) = \sin x$

  • $f(x) = x$

• Properties:

  • If f and g are odd, then f+g and f-g are odd, but f·g is even.

  • The composition of two odd functions is odd.

  • Any function can be written as the sum of an even and an odd function.

Domain and Range Calculations

23. The domain of the function $f(x) = \frac{1}{x-2}$ is:

1. All real numbers

2. All real numbers except 0

3. All real numbers except 2

4. All real numbers except -2

Show me the answer

Answer: 3. All real numbers except 2

Explanation:

• For rational functions, the domain excludes values that make the denominator zero.

• Here, denominator = $x - 2$

• Set denominator ≠ 0: $x - 2 ≠ 0$ ⇒ $x ≠ 2$

• Therefore, domain = $\{x \in \mathbb{R} : x ≠ 2\}$

• In interval notation: $(-\infty, 2) \cup (2, \infty)$

• The function has a vertical asymptote at x = 2.

24. The domain of the function $f(x) = \sqrt{x-3}$ is:

1. All real numbers

2. $x ≥ 0$

3. $x ≥ 3$

4. $x ≤ 3$

Show me the answer

Answer: 3. $x ≥ 3$

Explanation:

• For square root functions, the expression inside the square root must be non-negative.

• Here: $x - 3 ≥ 0$ ⇒ $x ≥ 3$

• Therefore, domain = $[3, \infty)$

• The range is $[0, \infty)$ because square root always gives non-negative outputs.

• Check: $f(3) = \sqrt{3-3} = \sqrt{0} = 0$ $f(4) = \sqrt{4-3} = \sqrt{1} = 1$ $f(7) = \sqrt{7-3} = \sqrt{4} = 2$

25. The domain of the function $f(x) = \log(x-1)$ is:

1. All real numbers

2. $x > 0$

3. $x > 1$

4. $x ≥ 1$

Show me the answer

Answer: 3. $x > 1$

Explanation:

• For logarithmic functions, the argument must be strictly positive.

• Here: $x - 1 > 0$ ⇒ $x > 1$

• Therefore, domain = $(1, \infty)$

• Common mistake: $x ≥ 1$ is incorrect because $\log(0)$ is undefined.

• The range of logarithmic functions is all real numbers: $(-\infty, \infty)$.

26. The range of the function $f(x) = x^2 + 4$ is:

1. All real numbers

2. $[0, \infty)$

3. $[4, \infty)$

4. $(-\infty, 4]$

Show me the answer

Answer: 3. $[4, \infty)$

Explanation:

• For $f(x) = x^2 + 4$:

  • $x^2 ≥ 0$ for all real x

  • Therefore, $x^2 + 4 ≥ 4$ for all real x

• The minimum value occurs at x = 0: $f(0) = 0^2 + 4 = 4$

• As $|x|$ increases, $f(x)$ increases without bound.

• Therefore, range = $[4, \infty)$

• Graphically, this is a parabola opening upward with vertex at (0, 4).

Composite Functions and Inverses

27. If $f(x) = 2x - 1$ and $g(x) = x^2 + 3$, then $(f \circ g)(x)$ is:

1. $2x^2 + 5$

2. $2x^2 + 2$

3. $4x^2 - 4x + 4$

4. $2x^2 + 5x - 1$

Show me the answer

Answer: 2. $2x^2 + 2$

Explanation:

• $(f \circ g)(x) = f(g(x)) = f(x^2 + 3)$

• Substitute $x^2 + 3$ into f: $2(x^2 + 3) - 1$

• Simplify: $2x^2 + 6 - 1 = 2x^2 + 5$

• Wait, let's recalculate carefully: $f(x) = 2x - 1$ $g(x) = x^2 + 3$ $f(g(x)) = f(x^2 + 3) = 2(x^2 + 3) - 1 = 2x^2 + 6 - 1 = 2x^2 + 5$

• Actually, option 1 is $2x^2 + 5$, not option 2. So the correct answer should be option 1.

• Let me verify: Option 1 says 2x² + 5, option 2 says 2x² + 2.

• Calculation shows 2x² + 5, so option 1 is correct.

Correction: The answer is 1. $2x^2 + 5$

28. If $f(x) = 2x - 1$ and $g(x) = x^2 + 3$, then $(g \circ f)(x)$ is:

1. $2x^2 + 5$

2. $4x^2 - 4x + 4$

3. $4x^2 - 4x + 2$

4. $2x^2 + 2$

Show me the answer

Answer: 2. $4x^2 - 4x + 4$

Explanation:

• $(g \circ f)(x) = g(f(x)) = g(2x - 1)$

• Substitute $2x - 1$ into g: $(2x - 1)^2 + 3$

• Expand: $(4x^2 - 4x + 1) + 3 = 4x^2 - 4x + 4$

• Therefore, $(g \circ f)(x) = 4x^2 - 4x + 4$

• Note that in general, $(f \circ g)(x) \neq (g \circ f)(x)$.

• Here: $(f \circ g)(x) = 2x^2 + 5$, while $(g \circ f)(x) = 4x^2 - 4x + 4$.

29. The inverse of $f(x) = \frac{2x+3}{x-1}$, for $x \neq 1$, is:

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

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

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

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

Show me the answer

Answer: 1. $f^{-1}(x) = \frac{x+3}{x-2}$

Explanation:

• To find inverse of $f(x) = \frac{2x+3}{x-1}$:

  1. Set $y = \frac{2x+3}{x-1}$

  2. Solve for x: $y(x-1) = 2x+3$ $yx - y = 2x + 3$ $yx - 2x = y + 3$ $x(y-2) = y+3$ $x = \frac{y+3}{y-2}$

  3. Swap x and y: $y = \frac{x+3}{x-2}$

• Therefore, $f^{-1}(x) = \frac{x+3}{x-2}$, for $x \neq 2$

• Verification: Check that $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$

30. If $f(x) = 3x + 2$ and $g(x) = \frac{x-2}{3}$, then:

1. g is the inverse of f

2. f is the inverse of g

3. Both 1 and 2

4. Neither is inverse of the other

Show me the answer

Answer: 3. Both 1 and 2

Explanation:

• Check if g is inverse of f: $(g \circ f)(x) = g(f(x)) = g(3x+2) = \frac{(3x+2)-2}{3} = \frac{3x}{3} = x$ $(f \circ g)(x) = f(g(x)) = f\left(\frac{x-2}{3}\right) = 3\left(\frac{x-2}{3}\right) + 2 = (x-2) + 2 = x$

• Both compositions give the identity function.

• Therefore, g = f⁻¹ and f = g⁻¹.

• This is symmetric: if g is the inverse of f, then f is also the inverse of g.

• The functions are mutual inverses.