Chapter 6 · Application of Derivatives

Miscellaneous · Q13: If $f'(x)>0$ on $(a,b)$ ⁠, then $f$ is increasing on $(a,b)$

Miscellaneous ExerciseIncreasing and decreasing functions · Section 6.3 · Miscellaneous Exercise

The question

Let $f$ be a function defined on $[a, b]$ such that $f'(x) > 0$ ⁠, for all $x \in (a, b)$ ⁠. Then prove that $f$ is an increasing function on $(a, b)$ ⁠.

Exam answer

Let $x_1, x_2 \in (a, b)$ with $x_1 < x_2$ ⁠. As $f'$ exists on $(a, b)$ ⁠, $f$ is continuous on $[x_1, x_2]$ and differentiable on $(x_1, x_2)$ ⁠.

By the Mean Value Theorem, there is $c \in (x_1, x_2)$ such that

f(x_2) - f(x_1) = f'(c)(x_2 - x_1)

Here $f'(c) > 0$ and $x_2 - x_1 > 0$ ⁠, so $f(x_2) - f(x_1) > 0$ ⁠, that is $f(x_1) < f(x_2)$ ⁠.

$\therefore\ x_1 < x_2 \Rightarrow f(x_1) < f(x_2)$ for all $x_1, x_2$ in $(a, b)$ ⁠, so $f$ is increasing on $(a, b)$ ⁠. Hence proved.

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

The one idea this problem is built on

The Mean Value Theorem turns information about the slope into information about values: between any two points the change in $f$ equals (slope at some point in between) $\times$ (change in $x$ ⁠). If every slope is positive, every such change is positive, so a point further right always has a higher value.

Given, to show, and the plan

Given: $f$ is defined on $[a, b]$ ⁠, and $f'(x)$ exists and is positive at every $x$ in the open interval $(a, b)$ ⁠. Nothing is said about $f$ at $a$ or $b$ beyond being defined there.
To show: $f$ is increasing on $(a, b)$ ⁠: for all $x_1, x_2 \in (a, b)$ ⁠, $x_1 < x_2 \Rightarrow f(x_1) \le f(x_2)$ (Definition 1 (i), Section 6.3).
Plan: (1) Pin down exactly what must be proved. (2) See what the hypothesis gives (differentiable, hence continuous, on $(a, b)$ ⁠). (3) Take arbitrary $x_1 < x_2$ in $(a, b)$ and apply the Mean Value Theorem on $[x_1, x_2]$ ⁠. (4) Read off the sign. (5) Conclude. (6) See why the result is stated on $(a, b)$ ⁠, not $[a, b]$ ⁠.

Basics you need

1The book's two kinds of "increasing" (Definition 1, Section 6.3)

On an interval I: increasing if $x_1 < x_2 \Rightarrow f(x_1) \le f(x_2)$ ⁠; strictly increasing if $x_1 < x_2 \Rightarrow f(x_1) < f(x_2)$ ⁠, for all $x_1, x_2$ in I. Strictly increasing implies increasing. Lesson: increasing and decreasing functions

2Differentiable implies continuous

If $f'(c)$ exists, then $f$ is continuous at $c$ (Theorem 3, Section 5.3). Basics: the derivative as a rate of change

3Mean Value Theorem

If $f$ is continuous on $[p, q]$ and differentiable on $(p, q)$ ⁠, there is a point $c$ with $p < c < q$ and $f(q) - f(p) = f'(c)(q - p)$ ⁠. The book uses it to prove Theorem 1 of Section 6.3, citing it as "Theorem 8 in Chapter 5"; that theorem is not printed in this reprint (its Chapter 5 ends at Section 5.7), so here it is quoted, not proved. Lesson: increasing and decreasing functions

4Sign of a product

The product of two positive numbers is positive. Basics: sign of a product

Solution

What exactly must be proved

By Definition 1 (i), " $f$ is increasing on $(a, b)$ " means
$x_1 < x_2 \text{ in } (a, b) \ \Rightarrow\ f(x_1) \le f(x_2).$
We shall prove the stronger statement $f(x_1) < f(x_2)$ (strictly increasing, Definition 1 (iv)); it implies the required one at once.

"For all $x_1, x_2$ " means the proof must work for an arbitrary pair, not for examples. And since the hypothesis is the strict inequality $f' > 0$ ⁠, we can expect, and will get, the strict conclusion.

If $f(x_1) < f(x_2)$ then certainly $f(x_1) \le f(x_2)$ ⁠. Lesson: increasing and decreasing
What the hypothesis gives us

$f'(x)$ exists at every $x \in (a, b)$ ⁠, so $f$ is differentiable on $(a, b)$ ⁠. A function is continuous wherever it is differentiable, so $f$ is also continuous at every point of $(a, b)$ ⁠.

The Mean Value Theorem needs continuity on a closed interval and differentiability inside it. Both will come from this step, as long as the closed interval sits inside $(a, b)$ ⁠. Note what we are not told: whether $f$ is continuous at the ends $a$ and $b$ ⁠.

Theorem 3, Section 5.3: differentiable at $c$ implies continuous at $c$ ⁠. Basics: the derivative
Apply the Mean Value Theorem on $[x_1, x_2]$

Let $x_1$ and $x_2$ be any two points of $(a, b)$ with $x_1 < x_2$ ⁠. Then
$a < x_1 < x_2 < b, \quad\text{so}\quad [x_1, x_2] \subset (a, b).$
By Step 2, $f$ is continuous on $[x_1, x_2]$ and differentiable on $(x_1, x_2)$ ⁠. The Mean Value Theorem gives a point $c$ with $x_1 < c < x_2$ and
$f(x_2)-f(x_1)=f'(c)(x_2-x_1).$
This is the heart of the proof. The closed interval $[x_1, x_2]$ lies strictly inside $(a, b)$ ⁠, so its end points $x_1$ ⁠, $x_2$ are points where $f$ is differentiable, hence continuous. That is why we never need anything about $f$ at $a$ or $b$ ⁠.

Mean Value Theorem (Basic 3; quoted, not printed in this reprint), used in the book's proof of Theorem 1, Section 6.3. Lesson: increasing and decreasing
Read off the sign

The point $c$ lies in $(x_1, x_2)$ ⁠, hence in $(a, b)$ ⁠, so by hypothesis $f'(c) > 0$ ⁠. Also $x_2 - x_1 > 0$ ⁠. Therefore
$f(x_2) - f(x_1) = \underbrace{f'(c)}_{>0}\,\underbrace{(x_2 - x_1)}_{>0} > 0,$
that is, $f(x_1) < f(x_2)$ ⁠.

We do not know where $c$ is, only that it lies between $x_1$ and $x_2$ ⁠. That is enough, because the hypothesis says $f'$ is positive at every point of $(a, b)$ ⁠, whichever one $c$ turns out to be.

Positive times positive is positive. Basics: sign of a product
Conclude

The pair $x_1 < x_2$ was arbitrary, so for all $x_1, x_2 \in (a, b)$ ⁠:
$x_1 < x_2 \ \Rightarrow\ f(x_1) < f(x_2).$
So $f$ is strictly increasing on $(a, b)$ (Definition 1 (iv)), and in particular $f$ is increasing on $(a, b)$ (Definition 1 (i)).

This is Theorem 1 (a) of Section 6.3 applied on every closed interval $[x_1, x_2]$ inside $(a, b)$ ⁠, with the strict inequality that the book's proof of Theorem 1 actually delivers.

Definition 1 (i), (iv), Section 6.3. Lesson: increasing and decreasing
Why the conclusion is about $(a, b)$ and not $[a, b]$

Theorem 1 (a) of Section 6.3 reads: 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 $x \in (a, b)$ ⁠. It needs continuity at $a$ and $b$ ⁠, which this question does not give. Without it the conclusion can fail on $[a, b]$ ⁠. For example, on $[0, 1]$ let
$g(0) = 1, \qquad g(x) = x \ \text{ for } 0 < x \le 1.$
Then $g'(x) = 1 > 0$ on $(0, 1)$ ⁠, and $g$ is increasing on $(0, 1)$ ⁠, as just proved. But $0 < \frac12$ while $g(0) = 1 > \frac12 = g\left(\frac12\right)$ ⁠, so $g$ is not increasing on $[0, 1]$ ⁠.

The function $g$ ⁠: the value at 0 is "out of line" with the rest (open dot at the origin, filled dot at $(0, 1)$ ⁠). The derivative is positive everywhere inside $(0, 1)$ ⁠, yet $g$ is not increasing on $[0, 1]$ ⁠, because $g$ is not continuous at 0.

This is why the question carefully asks only about $(a, b)$ ⁠. If, in addition, $f$ were continuous at $a$ and $b$ ⁠, Theorem 1 (a) would give "increasing on $[a, b]$ " as well.

A single counterexample is enough to show that a stronger claim is false. Lesson: increasing and decreasing

Result

For any $x_1 < x_2$ in $(a, b)$ ⁠, the Mean Value Theorem on $[x_1, x_2] \subset (a, b)$ gives $f(x_2) - f(x_1)$ $= f'(c)(x_2 - x_1) > 0$ ⁠. Hence $f$ is strictly increasing on $(a, b)$ ⁠, and therefore $f$ is an increasing function on $(a, b)$ . (Without continuity at $a$ and $b$ the same cannot be said of $[a, b]$ ⁠.) (The answer key in this reprint has no entry for this question.)

Check it with numbers

Take $f(x) = x^3 + x$ on $(-2, 2)$ ⁠: $f'(x) = 3x^2 + 1 > 0$ ⁠. On $[0, 1]$ the Mean Value point satisfies $f(1) - f(0) = 2 = f'(c)\cdot 1$ ⁠, so $3c^2 + 1 = 2$ ⁠, $c = \frac{1}{\sqrt3} \approx 0.5774$ ⁠, which does lie in $(0, 1)$ ⁠, and $f(0) = 0 < 2 = f(1)$ ✓. A computer check of 400 random points in $(-2, 2)$ ⁠, sorted, finds the values of $f$ strictly rising ✓.

Common mistakes

Applying Theorem 1 on $[a, b]$ directly. Its hypothesis "continuous on $[a, b]$ " is not given here. Apply the Mean Value Theorem on $[x_1, x_2]$ inside $(a, b)$ instead.
Choosing particular $x_1, x_2$ , such as the midpoint. The proof must work for every pair.
Assuming $c$ is a known point. The theorem only says $c$ exists between $x_1$ and $x_2$ ⁠; its exact position is never needed.
Proving only $f(x_1) \le f(x_2)$ and stopping is acceptable for "increasing", but the hypothesis gives the strict inequality, which is the more informative conclusion.

What to take away

"Positive derivative everywhere inside" gives "strictly increasing inside", by the Mean Value Theorem on an arbitrary sub-interval.
To extend the conclusion to the end points you need continuity there: compare the hypotheses of Theorem 1.
Every "find where $f$ is increasing" problem, such as Question 10's sign chart, rests on this argument.

Every fact used, and where it comes from

Fact	Used in	Source
Increasing ( $\le$ ⁠) and strictly increasing ( $<$ ⁠) on an interval	Steps 1, 5	Definition 1 (i), (iv), Section 6.3 (Lesson: increasing and decreasing)
Differentiable at $c$ implies continuous at $c$	Step 2	Theorem 3, Section 5.3 (Chapter 5)
Mean Value Theorem	Step 3	Quoted in the proof of Theorem 1, Section 6.3; the book cites it as "Theorem 8 in Chapter 5", but this reprint's Chapter 5 ends at Section 5.7 and does not contain it
Theorem 1 (a): continuous on $[a, b]$ ⁠, $f'(x) \ge 0$ on $(a, b)$ gives increasing in $[a, b]$	Steps 5, 6	Theorem 1, Section 6.3, as printed in this reprint
Counterexample and Mean Value point	Step 6, Check	Checked in the site's verification script

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Miscellaneous · Q13: If f′(x)>0f'(x)>0f′(x)>0 on (a,b)(a,b)(a,b)⁠, then fff is increasing on (a,b)(a,b)(a,b)

Basics you need

Solution

What exactly must be proved

What the hypothesis gives us

Apply the Mean Value Theorem on [x1,x2][x_1, x_2][x1​,x2​]

Read off the sign

Conclude

Why the conclusion is about (a,b)(a, b)(a,b) and not [a,b][a, b][a,b]

Every fact used, and where it comes from

Miscellaneous · Q13: If $f'(x)>0$ on $(a,b)$ ⁠, then $f$ is increasing on $(a,b)$

Apply the Mean Value Theorem on $[x_1, x_2]$

Why the conclusion is about $(a, b)$ and not $[a, b]$