Chapter 6 · Application of Derivatives

Example 7: $f(x) = 7x-3$ is increasing on $\R$

ExamplesIncreasing and decreasing functions · Section 6.3

The question

Show that the function given by $f(x) = 7x - 3$ is increasing on $\mathbf{R}$ ⁠.

Exam answer

$f(x) = 7x - 3$ ⁠, so $f'(x) = 7$ ⁠.

For all $x \in \mathbf{R}$ ⁠, $f'(x) = 7 > 0$ ⁠.

Hence $f$ is increasing (in fact strictly increasing) on $\mathbf{R}$ ⁠.

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

The one idea this problem is built on

"Increasing on $\mathbf{R}$ " is a statement about every pair of real numbers: whenever $x_1 < x_2$ ⁠, the values must satisfy $f(x_1) \le f(x_2)$ (Definition 1). To prove a "for every pair" statement, take two arbitrary numbers $x_1 < x_2$ and show the inequality for them. For $7x - 3$ the difference $f(x_2) - f(x_1)$ is simply $7(x_2 - x_1)$ ⁠, which is positive.

Given, to find, and the plan

Given: $f(x) = 7x - 3$ for every real $x$ ⁠.
To show: $f$ is increasing on $\mathbf{R}$ ⁠: for all $x_1, x_2 \in \mathbf{R}$ ⁠, $x_1 < x_2 \Rightarrow f(x_1) \le f(x_2)$ ⁠.
Plan: Take arbitrary $x_1 < x_2$ ⁠; compute $f(x_2) - f(x_1)$ ⁠; show it is positive. This proves even "strictly increasing", which implies "increasing". Then cross-check with the derivative (Theorem 1).

Basics you need

1Definition 1 (Section 6.3)

Let I be an interval in the domain of $f$ ⁠. Then $f$ is increasing on I if $x_1 < x_2$ in I $\Rightarrow f(x_1) \le f(x_2)$ ⁠, and strictly increasing on I if $x_1 < x_2$ in I $\Rightarrow f(x_1) < f(x_2)$ ⁠, for all $x_1, x_2 \in$ I. Lesson: increasing and decreasing functions

2Rules for inequalities

Multiplying both sides of an inequality by a positive number keeps its direction; adding or subtracting the same number on both sides keeps its direction. So $a < b \Rightarrow 7a < 7b$ $\Rightarrow 7a - 3 < 7b - 3$ ⁠.

3Theorem 1 (Section 6.3)

If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ ⁠, then $f$ is increasing in $[a, b]$ if $f'(x) \ge 0$ for each $x \in (a, b)$ ⁠. Lesson: increasing and decreasing functions

Solution

Take two arbitrary real numbers

Let $x_1$ and $x_2$ be any two real numbers with $x_1 < x_2$ ⁠. Nothing else is assumed about them.

The claim is about all pairs. Checking a few pairs, such as 0 and 1, can never prove it, because there are infinitely many pairs. An argument that uses only " $x_1 < x_2$ " works for every pair at once.

Definition 1, Section 6.3. Lesson: increasing and decreasing functions
Compare $f(x_1)$ and $f(x_2)$

Subtract:
$\begin{aligned} f(x_2) - f(x_1) &= (7x_2 - 3) - (7x_1 - 3) \\ &= 7(x_2-x_1) . \end{aligned}$
Since $x_1 < x_2$ ⁠, the bracket $x_2 - x_1$ is positive, and 7 is positive, so $f(x_2) - f(x_1) > 0$ ⁠, that is
$x_1 < x_2 \;\Longrightarrow\; f(x_1) < f(x_2) .$
Looking at the difference is the cleanest way to compare two numbers: a positive difference means the second is bigger. The $-3$ cancels, so only the coefficient 7 matters. The book reaches the same conclusion in one chain: $x_1 < x_2 \Rightarrow 7x_1 < 7x_2$ $\Rightarrow 7x_1 - 3 < 7x_2 - 3$ (Basic 2).

A product of two positive numbers is positive. Basics: sign of a product
Conclude with Definition 1

For all $x_1, x_2 \in \mathbf{R}$ ⁠, $x_1 < x_2$ gives $f(x_1) < f(x_2)$ ⁠. By Definition 1(iv), $f$ is strictly increasing on $\mathbf{R}$ . Since $f(x_1) < f(x_2)$ certainly gives $f(x_1) \le f(x_2)$ ⁠, $f$ is also increasing on $\mathbf{R}$ in the sense of Definition 1(i), which is what the question asks.

The graph of $y = 7x - 3$ (not to scale: one unit on the $y$ -axis is drawn much shorter than one unit on the $x$ -axis). For any $x_1 < x_2$ ⁠, the point above $x_2$ is higher.

The book distinguishes "increasing" ( $\le$ ⁠) from "strictly increasing" ( $<$ ⁠). The question asks for the weaker property; we proved the stronger one, which includes it. The book's solution also ends with "strictly increasing on $\mathbf{R}$ ".

Definition 1, parts (i) and (iv), Section 6.3. Lesson: increasing and decreasing functions
Cross-check with the derivative (Theorem 1)

$f'(x) = 7$ for every $x$ ⁠. Take any interval $[a, b]$ ⁠: $f$ is a polynomial, so it is continuous on $[a, b]$ and differentiable on $(a, b)$ ⁠, and $f'(x) = 7 \ge 0$ there. By Theorem 1(a), $f$ is increasing in $[a, b]$ ⁠. Any two numbers $x_1 < x_2$ lie in the interval $[x_1, x_2]$ ⁠, so $f(x_1) \le f(x_2)$ ⁠: increasing on $\mathbf{R}$ again.

Theorem 1 is the tool used for harder functions (from Example 8 on), where the difference $f(x_2) - f(x_1)$ cannot be simplified so easily. Here both routes are short, and they agree. The slope of the line is 7, positive, so the line rises everywhere.

Theorem 1, Section 6.3. Lesson: increasing and decreasing functions

Result

For any $x_1 < x_2$ ⁠, $f(x_2) - f(x_1) = 7(x_2 - x_1) > 0$ ⁠. Hence $f(x) = 7x - 3$ is strictly increasing on $\mathbf{R}$ , and so it is increasing on $\mathbf{R}$ ⁠. This agrees with the book's solution.

Check it with numbers

$x$	$-2$	$0$	$0.2$	$1$	$3$
$f(x)$	$-17$	$-3$	$-1.6$	4	18

The values rise from left to right, and each step equals 7 times the step in $x$ ⁠: for instance from $x = 0.2$ to $x = 1$ the value rises by $4 - (-1.6) = 5.6 = 7 \times 0.8$ ⁠. (The points $x_1 = 0.2$ and $x_2 = 1$ are the ones marked in the figure.)

Common mistakes

Proving with examples. " $f(0) = -3 < 4 = f(1)$ " checks one pair only; the definition needs every pair.
Mixing up the two definitions. Increasing uses $\le$ ⁠, strictly increasing uses $<$ ⁠. Strictly increasing implies increasing, not the other way round.
Reversing the inequality when multiplying. Multiplying by 7 keeps $<$ because 7 is positive; multiplying by a negative number would reverse it.

What to take away

To prove "increasing on I", take arbitrary $x_1 < x_2$ in I and compare $f(x_1)$ with $f(x_2)$ ⁠.
For $f(x) = mx + c$ ⁠, $f(x_2) - f(x_1) = m(x_2 - x_1)$ ⁠: the line is strictly increasing if $m > 0$ and strictly decreasing if $m < 0$ ⁠.
Definition 1 and Theorem 1 must agree; use whichever is shorter.

Every fact used, and where it comes from

Fact	Used in	Source
Increasing and strictly increasing on an interval	Steps 1, 3	Definition 1 (i), (iv), Section 6.3 (Lesson)
Multiplying by a positive number keeps an inequality	Step 2	Algebra of inequalities
Product of positive numbers is positive	Step 2	Basics: sign of a product
$f' \ge 0$ on $(a,b)$ gives increasing on $[a,b]$	Step 4	Theorem 1(a), Section 6.3
Table of values	Check	Arithmetic, recomputed in the site's verification script

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Example 7: f(x)=7x−3f(x) = 7x-3f(x)=7x−3 is increasing on R\RR

Basics you need

Solution

Take two arbitrary real numbers

Compare f(x1)f(x_1)f(x1​) and f(x2)f(x_2)f(x2​)

Conclude with Definition 1

Cross-check with the derivative (Theorem 1)

Every fact used, and where it comes from

Example 7: $f(x) = 7x-3$ is increasing on $\R$

Compare $f(x_1)$ and $f(x_2)$