Chapter 1 · Relations and Functions

Exercise 1.2 · Q11 (MCQ): $f(x)=x^4$ on $\R$

Exercise 1.2Types of functions · Exercise 1.2

The question

Let $f : \R \to \R$ be defined as $f(x) = x^4$ ⁠. Choose the correct answer.

(A) $f$ is one-one onto
(B) $f$ is many-one onto
(C) $f$ is one-one but not onto
(D) $f$ is neither one-one nor onto.

Exam answer

$f(1) = 1 = f(-1)$ with $1 \ne -1$ ⁠, so $f$ is not one-one. $x^4 \ge 0$ for all $x$ ⁠, so $-1$ has no preimage, and $f$ is not onto.

Answer: (D) $f$ is neither one-one nor onto

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

An even power such as $x^4$ has two features. It forgets the sign of $x$ ⁠, since $(-x)^4 = x^4$ ⁠, so different inputs share outputs: not one-one. And it is never negative, since $x^4 = (x^2)^2 \ge 0$ ⁠, so negative numbers in the codomain $\R$ are never reached: not onto. Once both properties are settled, exactly one option matches.

Given, to find, and the plan

Given: $f(x) = x^4$ ⁠, with domain $\R$ and codomain $\R$ ⁠.
To find: Which of (A)–(D) correctly describes $f$ ⁠.
Plan: (1) Decide one-one: look for a counterexample. (2) Decide onto: find a real number that $x^4$ never equals, and prove it. (3) Compare the two findings with each option and say why each wrong option fails.

Basics you need

1One-one, and "many-one"

$f$ is one-one if $f(x_1)=f(x_2) \Rightarrow x_1 = x_2$ for all $x_1, x_2$ ⁠. If this fails, the textbook calls $f$ many-one (Definition 5). So "many-one" simply means "not one-one". Basics: one-one and onto

2Onto

$f : X \to Y$ is onto if every $y\in Y$ equals $f(x)$ for some $x\in X$ (Definition 6). To disprove it, show one $y\in Y$ is never reached.

3Even powers

For every real $x$ ⁠: $(-x)^4 = x^4$ ⁠, and $x^4 = (x^2)^2 \ge 0$ ⁠, with equality only at $x = 0$ ⁠.

4Reading a multiple-choice question

Each option makes two claims, one about one-one and one about onto. An option is correct only if both its claims are true. Basics: proving and disproving

Solution

Not one-one: $1$ and $-1$ collide

Compute
$f(1) = 1^4 = 1 \qquad\text{and}\qquad f(-1) = (-1)^4 = 1.$
So $f(1) = f(-1)$ although $1 \ne -1$ ⁠. Hence $f$ is not one-one; it is many-one.

One-one is a "for all" claim, so a single pair of different inputs with the same output disproves it. The pair $\pm1$ is the simplest, but every $x \ne 0$ collides with $-x$ ⁠, because $(-x)^4 = \big((-x)^2\big)^2 = (x^2)^2 = x^4$ ⁠. For example $f(2) = f(-2) = 16$ ⁠.

$(-x)^4 = x^4$ (Basic 3). One counterexample disproves a "for all" statement. Basics: proving and disproving
Not onto: negative numbers are never reached

For every real $x$ ⁠, $x^4 = (x^2)^2 \ge 0$ ⁠, because the square of a real number is never negative. Now take the codomain element $-1 \in \R$ ⁠. If some real $x$ had $f(x) = -1$ ⁠, we would have $x^4 = -1 < 0$ ⁠, contradicting $x^4 \ge 0$ ⁠. So no real $x$ maps to $-1$ ⁠, and $f$ is not onto.

The range of $f$ is exactly $[0, \infty)$ ⁠: every $y \ge 0$ is reached, by $x = \sqrt[4]{y}$ ⁠, and no $y < 0$ is.

The graph of $y = x^4$ (horizontal and vertical scales differ). It is symmetric about the $y$ -axis, so the line $y = 1$ meets it twice, at $x = \pm1$ (not one-one). It never goes below the $x$ -axis, so the line $y = -1$ meets it nowhere (not onto).

To show a value is never reached we need a reason valid for every $x$ ⁠, not a list of trials. Writing $x^4$ as $(x^2)^2$ makes it a square, and squares are never negative. Any negative number would serve as the missing value; $-1$ is the simplest.

Onto fails if one codomain element has no preimage; equivalently, the range $[0,\infty)$ is smaller than the codomain $\R$ (Remark after Definition 6). Basics: one-one and onto

Examine every option

We have shown: $f$ is not one-one (Step 1) and not onto (Step 2). Now test each option's two claims.

Option	Claims	Verdict
(A)	one-one, onto	Wrong. Both claims fail: $f(1) = f(-1)$ ⁠, and $-1$ has no preimage.
(B)	many-one, onto	Wrong. "Many-one" is true, but "onto" is false, since $-1$ is never reached.
(C)	one-one, not onto	Wrong. "Not onto" is true, but "one-one" is false, since $f(1) = f(-1) = 1$ ⁠.
(D)	not one-one, not onto	Correct. Both claims are exactly what Steps 1 and 2 proved.

Each option is a pair of claims, and an option is right only when both are true. (B) and (C) each get one half right, which is why they are tempting. Checking every option, rather than stopping at the first plausible one, guards against such half-right choices.

Many-one means "not one-one" (Definition 5). Lesson: types of functions

Result

The correct answer is (D): $f(x) = x^4$ is neither one-one nor onto. It is not one-one because $f(1) = f(-1) = 1$ ⁠, and not onto because $x^4 \ge 0$ for all real $x$ ⁠, so $-1$ is never an output. (Textbook answer: D.)

Check it with numbers

Evaluate $f$ at $-3, -2, \dots, 3$ ⁠: the values are $81, 16, 1, 0, 1, 16, 81$ ⁠. Seven inputs give only four distinct outputs, so there are collisions ✓. The smallest value is $f(0) = 0$ ⁠; no input gives a negative value ✓.

Common mistakes

Choosing (B) because $x^4$ "takes every large value". It reaches every $y \ge 0$ ⁠, but the codomain is all of $\R$ ⁠, and negative values are missed.
Choosing (C) after looking only at $x \ge 0$ ⁠, where $x^4$ is one-one. The domain is all of $\R$ ⁠, and negative inputs copy the positive ones.
Not knowing what "many-one" means. In this textbook it means "not one-one" (Definition 5).

What to take away

Every even power $x^2, x^4, x^6, \dots$ from $\R$ to $\R$ is neither one-one nor onto, for the same two reasons: $(-x)^n = x^n$ and $x^n \ge 0$ ⁠.
In an MCQ about function types, settle one-one and onto separately first, then match.
Compare with Q12: the linear function $3x$ behaves the opposite way, being both one-one and onto.

Every fact used, and where it comes from

Fact	Used in	Source
$(-x)^4 = x^4$ ⁠; $x^4 = (x^2)^2 \ge 0$	Steps 1, 2	Properties of real numbers
Definition of one-one; "many-one" means not one-one	Steps 1, 3	Definition 5, Section 1.3 (Lesson: types of functions)
Definition of onto; onto ⇔ range = codomain	Step 2	Definition 6 and the Remark after it, Section 1.3
Same argument for $x^2$ on $\R$	Steps 1, 2	Example 11, Section 1.3
All numerical values	Check	Recomputed in the site's verification script

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Exercise 1.2 · Q11 (MCQ): f(x)=x4f(x)=x^4f(x)=x4 on R\RR

Basics you need

Solution

Not one-one: 111 and −1-1−1 collide

Not onto: negative numbers are never reached

Examine every option

Every fact used, and where it comes from

Exercise 1.2 · Q11 (MCQ): $f(x)=x^4$ on $\R$

Not one-one: $1$ and $-1$ collide