Chapter 6 · Application of Derivatives

Exercise 6.2 · Q1: Show that $3x+17$ is increasing on $\R$

Exercise 6.2Increasing and decreasing functions · Section 6.3

The question

Show that the function given by $f(x) = 3x + 17$ is increasing on $\R$ ⁠.

Exam answer

$f(x) = 3x + 17$ ⁠, so $f'(x) = 3$ ⁠.

$f'(x) = 3 > 0$ for all $x \in \R$ ⁠.

Hence $f$ is increasing on $\R$ ⁠.

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

The one idea this problem is built on

"Increasing" is a statement about every pair of inputs: a smaller input never gives a larger output. For a straight line the slope says it all. Here every step of 1 to the right raises the height by exactly 3, so a point further right is always higher. We prove this twice: once from the definition (the way the book does Example 7) and once from the derivative $f'(x) = 3 > 0$ (Theorem 1).

Given, to find, and the plan

Given: $f(x) = 3x + 17$ for every real $x$ ⁠.
To show: $f$ is increasing on $\R$ in the sense of Definition 1 (i): $x_1 < x_2 \Rightarrow f(x_1) \le f(x_2)$ ⁠.
Plan: (1) Read the book's definitions so we know exactly what to prove. (2) Prove the stronger statement $f(x_1) < f(x_2)$ directly from the inequality $x_1 < x_2$ ⁠. (3) Confirm with the derivative and Theorem 1. (4) Conclude.

Basics you need

1Increasing and strictly increasing (Definition 1)

On an interval I, $f$ is increasing if $x_1 < x_2$ in I always gives $f(x_1) \le f(x_2)$ ⁠, and strictly increasing if it always gives $f(x_1) < f(x_2)$ ⁠. Strictly increasing is the stronger property: it implies increasing. Lesson: increasing and decreasing functions

2Inequalities: what keeps them the right way round

Adding the same number to both sides keeps an inequality: $a < b \Rightarrow a + c < b + c$ ⁠. Multiplying both sides by a positive number keeps it: $a < b \Rightarrow 3a < 3b$ ⁠. (Multiplying by a negative number would reverse it.)

3The derivative test (Theorem 1)

If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ with $f'(x) \ge 0$ at every point of $(a, b)$ ⁠, then $f$ is increasing on $[a, b]$ ⁠. The book's proof (by the Mean Value Theorem) shows more: if $f'(x) > 0$ on $(a, b)$ ⁠, then $x_1 < x_2$ gives $f(x_1) < f(x_2)$ ⁠, so $f$ is strictly increasing. Basics: the derivative as a rate of change

4The real line as an interval

$\R = (-\infty, \infty)$ is itself an interval, so Definition 1 applies to it directly. Basics: interval notation

Solution

Pin down what "increasing" means in this book

Definition 1 in Section 6.3 gives two different words:
$\begin{aligned} &\text{increasing on I:} && x_1 < x_2 \Rightarrow f(x_1) \le f(x_2),\\ &\text{strictly increasing on I:} && x_1 < x_2 \Rightarrow f(x_1) < f(x_2), \end{aligned}$
each for all $x_1, x_2$ in I. The question asks for "increasing", the weaker one. We will prove the stronger property, strictly increasing, because it gives the weaker one at once: if $f(x_1) < f(x_2)$ ⁠, then certainly $f(x_1) \le f(x_2)$ ⁠.

The two words differ only in " $\le$ " versus " $<$ ". A constant function is increasing in the book's sense (all values equal) but not strictly increasing. Knowing which one is asked tells you what a complete proof must show.

"For all $x_1, x_2$ " means we must argue with two arbitrary numbers, not with examples. Lesson: increasing and decreasing functions
Proof from the definition

Let $x_1$ and $x_2$ be any two real numbers with $x_1 < x_2$ ⁠. Then
$\begin{aligned} x_1 < x_2 &\Rightarrow 3x_1 < 3x_2 \\ &\Rightarrow 3x_1 + 17 < 3x_2 + 17 \\ &\Rightarrow f(x_1) < f(x_2). \end{aligned}$
So $x_1 < x_2 \Rightarrow f(x_1) < f(x_2)$ for all $x_1, x_2 \in \R$ ⁠: by Definition 1 (iv), $f$ is strictly increasing on $\R$ , and therefore increasing on $\R$ ⁠.

The first implication multiplies both sides by 3. Because 3 is positive, the inequality keeps its direction. The second adds 17 to both sides, which never changes the direction. The last line just recognises $3x + 17$ as $f(x)$ ⁠. This is exactly the argument of Example 7 ( $7x - 3$ ⁠), with 3 and 17 in place of 7 and $-3$ ⁠.

Inequality rules: $a < b$ and $c > 0$ give $ca < cb$ ⁠; $a < b$ gives $a + d < b + d$ for any $d$ ⁠.
The same conclusion from the derivative

$f$ is a polynomial, so it is continuous and differentiable at every real number, and
$f'(x) = \frac{d}{dx}(3x + 17) = 3 \quad\text{for every } x \in \R.$
Take any $x_1 < x_2$ ⁠. On the closed interval $[x_1, x_2]$ ⁠, $f$ is continuous, it is differentiable on $(x_1, x_2)$ ⁠, and $f'(x) = 3 > 0$ there. By Theorem 1 (a) and its proof, $f(x_1) < f(x_2)$ ⁠. In numbers, the Mean Value step of the proof reads
$f(x_2) - f(x_1) = 3(x_2 - x_1) > 0.$
For a straight line this needs no special point $c$ ⁠. Directly,
$\begin{aligned} f(x_2) - f(x_1) &= (3x_2 + 17) - (3x_1 + 17)\\ &= 3(x_2 - x_1). \end{aligned}$
Theorem 1 is stated on a closed interval $[a, b]$ ⁠, while the question is about the whole line $\R$ ⁠. The trick is that any two numbers $x_1 < x_2$ lie in the closed interval $[x_1, x_2]$ ⁠, so applying the theorem there for every pair covers all of $\R$ ⁠.

The derivative of $ax + b$ is the constant $a$ ⁠: the slope of the line. A positive slope means the line rises from left to right. Basics: the derivative as a rate of change
Look at the picture, then conclude

The line $y = 3x + 17$ has slope 3. With $x_1 = 1$ and $x_2 = 3$ ⁠: $f(x_1) = 20 < 26 = f(x_2)$ ⁠. Any point further right is higher.

Both proofs give $x_1 < x_2 \Rightarrow f(x_1) < f(x_2)$ on $\R$ ⁠. Hence $f(x) = 3x + 17$ is increasing on $\R$ , and in fact strictly increasing.

The picture is not a proof (it shows only part of the line), but it shows what the algebra says: the graph climbs steadily, 3 units up for every 1 unit across.

Result

For all real $x_1 < x_2$ ⁠, we have $f(x_2) - f(x_1) = 3(x_2 - x_1)$ ⁠, which is positive. So $f(x) = 3x + 17$ is strictly increasing on $\R$ ⁠, and therefore increasing on $\R$ ⁠, as required. (The answer key gives no answer for this "show that" question.)

Check it with numbers

$f(-4) = 5$ ⁠, $f(-2) = 11$ ⁠, $f(0) = 17$ ⁠, $f(1) = 20$ ⁠, $f(3) = 26$ ⁠, $f(10) = 47$ ⁠. The inputs increase and so do the outputs. Between $x = 1$ and $x = 3$ ⁠: $f(3) - f(1) = 6 = 3 \times 2$ ⁠, exactly slope times the step, as Step 3 says.

Common mistakes

Checking a few values and stopping. $f(1) < f(3)$ is one pair; "increasing on $\R$ " is about every pair. Use arbitrary $x_1 < x_2$ ⁠.
Applying Theorem 1 directly to $\R$ ⁠. The theorem is stated for a closed interval $[a, b]$ ⁠. Apply it on $[x_1, x_2]$ for an arbitrary pair, as in Step 3.
Mixing up the two words. In this book "increasing" allows equal values ( $\le$ ⁠); "strictly increasing" does not. Proving the strict version is fine, as long as you say it implies the one asked.

What to take away

A linear function $ax + b$ with $a > 0$ is strictly increasing on $\R$ ⁠; with $a < 0$ it is strictly decreasing.
Two routes to "increasing": the definition (manipulate $x_1 < x_2$ ⁠) or the sign of $f'$ (Theorem 1).
To use a closed-interval theorem on all of $\R$ ⁠, apply it on $[x_1, x_2]$ for arbitrary $x_1 < x_2$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
Increasing ( $\le$ ⁠) and strictly increasing ( $<$ ⁠) on an interval	Steps 1, 2	Definition 1 (i), (iv), Section 6.3 (Lesson: increasing and decreasing functions)
Same argument for $7x - 3$	Step 2	Example 7, Section 6.3
Multiplying by a positive number and adding a number keep an inequality	Step 2	Rules for inequalities
$\frac{d}{dx}(3x + 17) = 3$	Step 3	Algebra of derivatives and $\frac{d}{dx}x^n = nx^{n-1}$ ⁠, Section 5.3 and Table 5.3
$f' \ge 0$ on $(a, b)$ gives increasing on $[a, b]$ ⁠; its proof gives strict increase when $f' > 0$	Step 3	Theorem 1 (a) and its proof, Section 6.3

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Exercise 6.2 · Q1: Show that 3x+173x+173x+17 is increasing on R\RR

Basics you need

Solution

Pin down what "increasing" means in this book

Proof from the definition

The same conclusion from the derivative

Look at the picture, then conclude

Every fact used, and where it comes from

Exercise 6.2 · Q1: Show that $3x+17$ is increasing on $\R$