Single-Variable-Calculus · Unit 9 · Video 3 · Interactive Practice

x / ln x: Inflection Point and Where Max and Min Can Occur

IKey Formulas

Formula	Name	What it tells you
$f'(x) = \dfrac{\ln x - 1}{(\ln x)^2} = \dfrac{1}{\ln x} - \dfrac{1}{(\ln x)^2}$	Split form of $f'$ for $f(x) = \dfrac{x}{\ln x}$	Constant numerators: both terms tend to $0$ as $\ln x \to -\infty$ , so $f'(0^+) = 0$
$f''(x) = \dfrac{2 - \ln x}{x(\ln x)^3}$	Second derivative (chain rule on $(\ln x)^{-1} - (\ln x)^{-2}$ )	Its sign can change only at $x = e^2$ (numerator) and $x = 1$ (denominator)
$\left(e^2,\ \tfrac{e^2}{2}\right) \approx (7.4,\ 3.7)$	Inflection point	Concave up on $(1, e^2)$ , concave down for $x > e^2$
critical points, endpoints, discontinuities	Where a max or min can occur	Check all three kinds, not only $f'(x) = 0$

Key Insight: For $x/\ln x$ there are five places to look ( $x \to 0^+$ , $x \to 1^-$ , $x \to 1^+$ , $x \to \infty$ and $x = e$ ) and only one of them is a critical point; over the whole domain there is no maximum and no minimum.

IIVisualization 1 — The Slope Where the Curve Leaves the Origin

In the split form both numerators are constant, so $f'(x) \to 0$ as $\ln x \to -\infty$ .

x = 10^(−k), k 0.25

IIIVisualization 2 — Concavity from the Sign of $f''$

The sign of $f''$ flips at $x = 1$ and at $x = e^2$ , but only $x = e^2$ is an inflection point.

x 4

IVVisualization 3 — Five Places to Look

A max or min can occur only at a critical point, an endpoint or a discontinuity, so every one must be checked.

Domain a 2 b 10

VQuiz Questions

Problem 1 · Where Is It Concave Up?

Given: $g(x) = xe^{-x}$ , with $g'(x) = (1 - x)e^{-x}$ and $g''(x) = (x - 2)e^{-x}$ — find where $g$ is concave up.

✅ Correct! Since

e^{-x} > 0

, the sign of

g''

is the sign of

x - 2

: positive exactly when

x > 2

❌ Check the sign. For

x < 2

the factor

x - 2

is negative, so

g'' < 0

there: that is where

g

is concave down.

❌ That is the turning point.

x = 1

is where

g' = 0

. The bend changes where

g''

changes sign, at a different place, just as

x/\ln x

turns at

e

but bends at

e^2

❌ Not quite.

g''(0) = -2 < 0

, so

g

is not concave up everywhere. Read the sign of

g''(x) = (x - 2)e^{-x}

Show solution

The factor $e^{-x}$ is positive for every $x$ , so

\operatorname{sign} g''(x) = \operatorname{sign}(x - 2).

For $x < 2$ : $g'' < 0$ , concave down. For $x > 2$ : $g'' > 0$ , concave up. The inflection point is $\left(2, \tfrac{2}{e^2}\right)$ , to the right of the turning point $x = 1$ found from $g' = 0$ .

Problem 2 · A Sign Change That Is Not an Inflection Point

Given: $h(x) = \dfrac{x^2}{x - 1}$ , with $h'(x) = \dfrac{x(x - 2)}{(x - 1)^2}$ and $h''(x) = \dfrac{2}{(x - 1)^3}$ — find the inflection points of $h$ .

✅ Correct!

h

is concave down for

x < 1

and concave up for

x > 1

, but

x = 1

is not in the domain, so there is no point of the graph where the bend changes.

❌ Not a point of the graph.

h(1)

is undefined:

x = 1

is a vertical asymptote, exactly like

x = 1

for

x/\ln x

, where

f''

also changes sign.

❌ That is a critical point.

h'(2) = 0

gives the turning point

(2, 4)

. An inflection point needs

h''

to change sign, and

h''(x) = 2/(x - 1)^3

keeps one sign on each side of

x = 1

❌ Right count, wrong concavity. For

x < 1

the cube

(x - 1)^3

is negative, so

h'' < 0

and

h

is concave down there.

Show solution

The numerator of $h''$ is the constant $2$ , so the sign of $h''$ is the sign of $(x - 1)^3$ :

x < 1:\ h'' < 0 \ \text{(concave down)}, \qquad x > 1:\ h'' > 0 \ \text{(concave up)}.

The concavity changes only across $x = 1$ , where $h$ is undefined (a vertical asymptote). An inflection point must be a point on the graph, so $h$ has no inflection points. The point $x = 2$ is a critical point, not a change of bend.

Problem 3 · The Inflection Point of $\ln x / x$

Given: $k(x) = \dfrac{\ln x}{x}$ for $x > 0$ , with $k'(x) = \dfrac{1 - \ln x}{x^2}$ — find the inflection point of $k$ .

What is its $x$ -coordinate?

What is its $y$ -coordinate?

✅ Correct!

k''(x) = \dfrac{2\ln x - 3}{x^3}

changes sign at

\ln x = \tfrac{3}{2}

, giving the inflection point

\left(e^{3/2}, \tfrac{3}{2e^{3/2}}\right) \approx (4.48,\ 0.335)

❌ That is the critical point.

k'(e) = 0

because

1 - \ln e = 0

: the maximum

(e, 1/e)

. The inflection point comes from

k''

❌ That belongs to $x/\ln x$ . Differentiate

k'(x) = (1 - \ln x)x^{-2}

yourself; the numerator of

k''

is not

2 - \ln x

❌ Check the $x$ -coordinate. Use the product rule on

(1 - \ln x)\,x^{-2}

, put both terms over

x^3

, and solve numerator

= 0

❌ Check the $y$ -coordinate. Evaluate

k

x = e^{3/2}

\ln\left(e^{3/2}\right) = \tfrac{3}{2}

, divided by

e^{3/2}

Show solution

Step 1: Differentiate again. Write $k'(x) = (1 - \ln x)\,x^{-2}$ and use the product rule:

k''(x) = -\frac{1}{x}\cdot x^{-2} + (1 - \ln x)(-2x^{-3}) = \frac{-1 - 2 + 2\ln x}{x^3} = \frac{2\ln x - 3}{x^3}.

Step 2: Sign. The denominator $x^3$ is positive for $x > 0$ , so the sign changes only where $2\ln x - 3 = 0$ , i.e. $\ln x = \tfrac{3}{2}$ , $x = e^{3/2} \approx 4.48$ : concave down before, concave up after.

Step 3: Height.

k\left(e^{3/2}\right) = \frac{3/2}{e^{3/2}} = \frac{3}{2e^{3/2}} \approx 0.335.

The inflection point $\left(e^{3/2}, \tfrac{3}{2e^{3/2}}\right)$ lies to the right of the maximum at $x = e$ .

Problem 4 · Checking Every Candidate

Given: $f(x) = x^2 + \dfrac{16}{x}$ , with $f'(x) = 2x - \dfrac{16}{x^2} = \dfrac{2(x^3 - 8)}{x^2}$ and the single critical point $x = 2$ — find the maximum on $[1, 4]$ and the minimum over the whole domain $x \neq 0$ .

Maximum value on $[1, 4]$ :

Minimum value over the whole domain $x \neq 0$ :

✅ Correct! On

[1, 4]

the maximum is at an endpoint, and over the whole domain the discontinuity at

x = 0

rules out a minimum. The critical point

x = 2

alone would have given neither answer.

❌ Compare both endpoints.

f(1) = 17

, but

f(4) = 16 + 4 = 20

is larger.

❌ That is the smallest value.

f(2) = 12

is the minimum on the interval; both endpoint values are larger.

❌ Only on $x > 0$ .

f(2) = 12

is the lowest value on the right branch, but for

x < 0

the term

\tfrac{16}{x}

is negative, and it drops to

-\infty

x \to 0^-

❌ $x = -2$ is not a candidate.

f'(-2) = \tfrac{2(-8 - 8)}{4} = -8 \neq 0

. On

x < 0

we have

f' < 0

, so

f

keeps falling, down to

-\infty

x \to 0^-

❌ Check the maximum.

f

is continuous on

[1, 4]

; compare the values at

x = 1

x = 2

and

x = 4

❌ Check the minimum.

f

does have a critical point,

x = 2

. Look at the discontinuity

x = 0

as well.

Show solution

On $[1, 4]$ the candidates are the critical point $x = 2$ and the two endpoints; there is no discontinuity in the interval.

f(1) = 1 + 16 = 17, \qquad f(2) = 4 + 8 = 12, \qquad f(4) = 16 + 4 = 20.

Maximum $20$ at the endpoint $x = 4$ ; minimum $12$ at the critical point $x = 2$ .

Over $x \neq 0$ the candidates are $x = 2$ , the ends $x \to \pm\infty$ and both sides of the discontinuity at $x = 0$ :

\lim_{x \to 0^-}\left(x^2 + \frac{16}{x}\right) = -\infty, \qquad \lim_{x \to 0^+}\left(x^2 + \frac{16}{x}\right) = +\infty.

So $f$ has no minimum (and no maximum). The critical value $12$ is only a local minimum.

Solved: 0 / 4

x / ln x: Inflection Point and Where Max and Min Can Occur

IKey Formulas

IIVisualization 1 — The Slope Where the Curve Leaves the Origin

IIIVisualization 2 — Concavity from the Sign of f′′f''f′′

IVVisualization 3 — Five Places to Look

VQuiz Questions

IIIVisualization 2 — Concavity from the Sign of $f''$