Single-Variable-Calculus · Unit 11 · Video 5 · Interactive Practice

Consequences of the Mean Value Theorem: Increasing Functions and Proving Inequalities

IKey Formulas

Formula	Name	What it gives
$f(b) = f(a) + f'(c)(b - a),\ \ a < c < b$	Mean Value Theorem, rearranged	Exact; $c$ is unknown
$f' > 0 \Rightarrow f$ increasing; $\ f' < 0 \Rightarrow f$ decreasing; $\ f' = 0 \Rightarrow f$ constant	Sign consequences	Sign of $f'(c)$ times $b - a > 0$
$\displaystyle \min_{a \le x \le b} f'(x) \le \frac{f(b) - f(a)}{b - a} \le \max_{a \le x \le b} f'(x)$	Mean Value bound	Average rate between extreme rates
$f(x) \approx f(a) + f'(a)(x - a)$	Linear approximation	Approximate, for $x$ near $a$

Key Insight: The derivative is a limit, a statement about infinitesimal change; "increasing" compares $f(b)$ with $f(a)$ a finite distance apart. The Mean Value Theorem is the exact bridge between the two, at the price of not knowing $c$ .

IIVisualization 1 — Exact Slope, Unknown Point

For $f(x) = \tfrac{x^3}{9} - x + 2$ , the line from $(a, f(a))$ with slope $f'(c)$ lands exactly on $(b, f(b))$ .

IIIVisualization 2 — The Average Rate Is Trapped

The unknown slope $f'(c)$ always lies between the smallest and largest values of $f'$ on $[a, b]$ , endpoints included.

💡 Challenge: shrink $[a, b]$ around $a = 0$ : the bounds squeeze together and the average rate approaches the initial speed $f'(a)$ , which is when linear approximation is accurate.

IVVisualization 3 — The Inequality Ladder for $e^x$

Each difference $h_n(x) = e^x - P_n(x)$ starts at $0$ and, for $x > 0$ , has derivative $h_{n-1}(x) > 0$ , so it stays positive there.

Step 1 x 1.50

Step 1 — $e^x > 1 + x$

h_1(x) = e^x - (1 + x), \qquad h_1(0) = 1 - 1 = 0

h_1'(x) = e^x - 1 > 0 \quad (x > 0)

e^x - 1

is itself

0

0

with derivative

e^x > 0

; so

h_1

is increasing and

h_1(x) > h_1(0) = 0

Step 2 — $e^x > 1 + x + \frac{x^2}{2}$

h_2(x) = e^x - \left(1 + x + \tfrac{x^2}{2}\right), \qquad h_2(0) = 0

h_2'(x) = e^x - (0 + 1 + x) = h_1(x) > 0

Step 3 — add $\frac{x^3}{3!}$ , with $3! = 6$

\left(\tfrac{x^3}{6}\right)' = \tfrac{3x^2}{6} = \tfrac{x^2}{2}

h_3'(x) = h_2(x) > 0 \ \Rightarrow\ e^x > 1 + x + \tfrac{x^2}{2} + \tfrac{x^3}{6}

Step 4 — add $\frac{x^4}{4!}$ , with $4! = 24$

\left(\tfrac{x^4}{24}\right)' = \tfrac{x^3}{6}, \qquad h_4'(x) = h_3(x) > 0

e^x > 1 + x + \tfrac{x^2}{2} + \tfrac{x^3}{3!} + \tfrac{x^4}{4!} \quad (x > 0)

💡 In the limit the tortoise catches the hare: $e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \cdots$ , the Taylor series of $e^x$ .

VQuiz Questions

Problem 1 · Read the Rearranged Theorem

Given: $f$ is differentiable everywhere, $f(2) = 5$ , and $f'(x) \ge 3$ for all $x$ . Find the smallest value $f(6)$ could possibly have.

✅ Correct!

f(6) = 5 + f'(c)\cdot 4 \ge 5 + 3 \cdot 4 = 17

, and

f(x) = 3x - 1

attains it.

❌ Not quite. The slope multiplies the length

b - a = 4

: the term is

f'(c)(b - a)

, not

f'(c)

❌ Not quite.

f'(c)(b - a) \ge 12

is the change

f(6) - f(2)

; add the starting value

f(2) = 5

❌ Not quite. The factor is the length

b - a = 6 - 2 = 4

, not

b = 6

Show solution

Apply the rearranged theorem with $a = 2$ , $b = 6$ :

f(6) = f(2) + f'(c)(6 - 2) = 5 + 4f'(c), \qquad 2 < c < 6.

Since $f'(c) \ge 3$ and $4 > 0$ , $f(6) \ge 5 + 12 = 17$ . The value is reached by $f(x) = 3x - 1$ , so the least possible value is $\mathbf{17}$ .

Problem 2 · Exact, Not Approximate

Given: $f(x) = \sqrt{x}$ on $[100, 101]$ , so $f'(x) = \frac{1}{2\sqrt{x}}$ . Which statement does the Mean Value Theorem prove?

✅ Correct!

f'(c) = \frac{1}{2\sqrt{c}} < \frac{1}{20}

because

c > 100

; indeed

\sqrt{101} \approx 10.0499

❌ Not quite.

10 + \frac{1}{20}

uses the slope

f'(100)

at the left end: that is the linear approximation. The theorem's slope is

f'(c)

at an unknown

c > 100

, so equality is not guaranteed (and is false).

❌ Check the direction.

f'

is decreasing, so

f'(c) < f'(100) = \frac{1}{20}

for every

c > 100

❌ Not quite.

f'(100) = \frac{1}{2 \cdot 10} = 0.05

, not

0.005

Show solution

With $a = 100$ , $b = 101$ :

\sqrt{101} = \sqrt{100} + \frac{1}{2\sqrt{c}}(101 - 100) = 10 + \frac{1}{2\sqrt{c}}, \qquad 100 < c < 101.

Because $c > 100$ , $\sqrt{c} > 10$ and $\frac{1}{2\sqrt{c}} < \frac{1}{20} = 0.05$ . Hence $\sqrt{101} < 10.05$ . The tangent-line value $10.05$ is only an approximation, and here an overestimate.

Problem 3 · Prove an Inequality by a Difference

Given: $h(x) = x - \ln(1 + x)$ for $x \ge 0$ , with $h(0) = 0$ . Find $h'(x)$ , then the inequality the method proves for $x > 0$ .

What is $h'(x)$ ?

Which inequality follows for $x > 0$ ?

✅ Correct!

h(0) = 0

and

h' > 0

for

x > 0

, so

h(x) > 0

, which says

\ln(1 + x) < x

❌ Check the derivative.

\frac{d}{dx}\ln(1 + x) = \frac{1}{1 + x}

, so

h'(x) = 1 - \frac{1}{1 + x}

; combine over a common denominator.

❌ Check the direction.

h(x) = x - \ln(1 + x) > 0

puts

x

on top:

x > \ln(1 + x)

❌ Not quite. The method only shows

h(x) > 0

; read that off directly. That bound is false: at

x = 1

its right-hand side is

\frac12

, but

\ln 2 \approx 0.693

Show solution

Derivative:

h'(x) = 1 - \frac{1}{1 + x} = \frac{(1 + x) - 1}{1 + x} = \frac{x}{1 + x}.

For $x > 0$ both numerator and denominator are positive, so $h'(x) > 0$ .

Conclusion: $h$ is increasing on $[0, \infty)$ , so for $x > 0$ , $h(x) > h(0) = 0$ :

x - \ln(1 + x) > 0 \iff \ln(1 + x) < x.

Problem 4 · When the Sign Rule Seems to Fail

Given: $f(x) = \frac{1}{x}$ has $f'(x) = -\frac{1}{x^2} < 0$ at every point of its domain, yet $f(-1) = -1 < 1 = f(1)$ . Which statement explains this?

✅ Correct! The proof applies the theorem to the pair

a = -1

b = 1

, which needs

f

on the whole interval between them.

f

is decreasing on

(-\infty, 0)

and on

(0, \infty)

separately.

❌ Not quite.

f'(c) = -\frac{1}{c^2}

is negative at every

c \ne 0

, and

c = 0

is not in the domain.

❌ Not quite. That describes linear approximation. The Mean Value Theorem is exact on any interval, short or long, where its hypotheses hold.

❌ Not quite. The proof uses only the rearranged theorem, valid for any function continuous on

[a, b]

and differentiable on

(a, b)

Show solution

The proof of " $f' < 0 \Rightarrow$ decreasing" takes $a < b$ and writes $f(b) = f(a) + f'(c)(b - a)$ . That step needs $f$ continuous on $[a, b]$ and differentiable on $(a, b)$ .

For $a = -1$ , $b = 1$ the interval contains $0$ , where $f(x) = \frac{1}{x}$ is not defined, so the theorem does not apply. On any interval avoiding $0$ it does apply, so $f$ is decreasing on $(-\infty, 0)$ and on $(0, \infty)$ , though not on its whole domain.

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Consequences of the Mean Value Theorem: Increasing Functions and Proving Inequalities

IKey Formulas

IIVisualization 1 — Exact Slope, Unknown Point

IIIVisualization 2 — The Average Rate Is Trapped

IVVisualization 3 — The Inequality Ladder for exe^xex

VQuiz Questions

IVVisualization 3 — The Inequality Ladder for $e^x$