Exercise 1.2 · Q7: $3-4x$ and $1+x^2$ on $\R$

Basics you need

Part (i): $f(x) = 3 - 4x$

1. One-one: equal outputs force equal inputs

   Let $x_1, x_2$ be any real numbers with $f(x_1) = f(x_2)$⁠. Then

   $$\begin{aligned} 3 - 4x_1 &= 3 - 4x_2 \\ -4x_1 &= -4x_2 &&\text{(subtract 3 from both sides)}\\ x_1 &= x_2 &&\text{(divide both sides by } -4\text{)}. \end{aligned}$$

   So $f(x_1) = f(x_2)$ always forces $x_1 = x_2$⁠: $f$ is one-one.

   One-one is a "for all" claim, so we must work with arbitrary $x_1, x_2$⁠, not chosen numbers. The method is to undo the formula one operation at a time. Each step is reversible: subtracting 3 is always allowed, and dividing by $-4$ is allowed because $-4 \ne 0$⁠. That non-zero coefficient is the whole reason the function is one-one: a linear function $ax+b$ is one-one exactly when $a\ne0$⁠.
   Dividing both sides of an equation by a non-zero number keeps it equivalent (Basic 4). One-one: $f(x_1)=f(x_2)\Rightarrow x_1=x_2$ (Definition 5).

2. Onto: solve $3 - 4x = y$ for any real $y$

   Take an arbitrary $y\in\R$⁠. We look for a real $x$ with $f(x) = y$⁠:

   $$3 - 4x = y \ \Longrightarrow\ -4x = y - 3 \ \Longrightarrow\ x = \frac{y-3}{-4} = \frac{3-y}{4}.$$

   This $x$ is a real number for every real $y$ (the only division is by 4), so it lies in the domain $\R$⁠. Check that it works:

   $$f\!\left(\frac{3-y}{4}\right) = 3 - 4\cdot\frac{3-y}{4} = 3 - (3 - y) = y. \ ✓$$

   So every real $y$ is the image of $\frac{3-y}{4}$⁠: $f$ is onto.

   Onto says every element of the codomain is hit, so again we start from an arbitrary $y$⁠. Solving the equation tells us which $x$ to try; the check then proves that this $x$ really maps to $y$⁠. We must also make sure the $x$ we found belongs to the domain. Here the domain is all of $\R$ and $\frac{3-y}{4}$ is always real, so there is nothing to exclude.
   Onto: for an arbitrary $y$ in the codomain, find $x$ in the domain with $f(x)=y$⁠, then check (Definition 6). Basics: one-one and onto

3. Conclude for part (i), and picture it

   $f$ is one-one (Step 1) and onto (Step 2), so $f(x) = 3 - 4x$ is bijective.

   The graph gives a picture of both properties at once. A horizontal line at height $y$ collects all inputs with output $y$⁠. If no horizontal line meets the graph twice, no two inputs share an output (one-one); if every horizontal line meets the graph, every $y$ is an output (onto).
   Bijective = one-one and onto (Definition 7). Lesson: types of functions

Part (ii): $f(x) = 1 + x^2$

1. Not one-one: $1$ and $-1$ give the same output

   Compute

   $$f(1) = 1 + 1^2 = 2 \qquad\text{and}\qquad f(-1) = 1 + (-1)^2 = 1 + 1 = 2.$$

   So $f(1) = f(-1)$ although $1 \ne -1$⁠. Two different inputs share an output, so $f$ is not one-one.

   Squaring forgets the sign: $(-x)^2 = x^2$⁠, so $f(-x) = f(x)$ for every $x$⁠. Any non-zero $x$ and its negative collide. One such pair is enough, because one-one is a "for all" claim and a single counterexample disproves it. Compare part (i): there, dividing by $-4$ undid the formula; here, the step $x_1^2 = x_2^2$ cannot be undone to $x_1 = x_2$⁠, only to $x_1 = \pm x_2$⁠.
   $(-x)^2 = x^2$ (Basic 5). To disprove a "for all" statement, one counterexample is enough. Basics: proving and disproving

2. Not onto: the output is never less than 1

   For every real $x$⁠, $x^2 \ge 0$ (Basic 5), so

   $$f(x) = 1 + x^2 \ \ge\ 1 + 0 = 1 \qquad\text{for all } x\in\R.$$

   Now take the codomain element $0\in\R$⁠. If some real $x$ had $f(x) = 0$⁠, then $1 + x^2 = 0$⁠, so $x^2 = -1$⁠, which is impossible for a real number because $x^2 \ge 0$⁠. So no $x\in\R$ maps to 0, and $f$ is not onto.

   The same holds for every $y < 1$⁠, for example $y = \tfrac12$ or $y = -5$⁠. The range of $f$ is exactly $[1, \infty)$⁠: every $y \ge 1$ is reached, by $x = \sqrt{y-1}$⁠, since $1 + (\sqrt{y-1})^2 = 1 + (y-1) = y$⁠.

   To show a value is never reached we need a reason that covers every $x$ at once; trying several inputs proves nothing. The inequality $x^2\ge0$ supplies that reason. We chose 0 as the missing value because it is the simplest number below 1, but any $y<1$ would do. Finding the full range $[1,\infty)$ shows precisely which part of the codomain is missed.
   Onto fails if some $y$ in the codomain has no preimage; equivalently, the range is smaller than the codomain (Remark after Definition 6). Basics: one-one and onto

3. Conclude for part (ii)

   $f$ is not one-one (Step 1) and not onto (Step 2), so $f(x) = 1 + x^2$ is neither one-one nor onto, and in particular not bijective.

   Both failures have a structural cause: squaring loses the sign (destroying one-one) and squaring never produces a negative number (destroying onto). Changing the domain to $[0,\infty)$ would repair one-one, and changing the codomain to $[1,\infty)$ would repair onto; with both changes $f$ would be a bijection. The question fixes both as $\R$⁠, so neither repair applies.

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