Single-Variable-Calculus · Unit 9 · Video 2 · Interactive Practice

A Five-Step Curve-Sketching Scheme, Applied to x / ln x

IKey Formulas

Formula	Step	Where it comes from
$\ln x = 0 \iff x = 1$ : $\ f(1^-) = \dfrac{1}{0^-} = -\infty$ , $\ f(1^+) = \dfrac{1}{0^+} = +\infty$	1(a) Discontinuities	Sign of $\ln x$ on each side of $1$
$f(0^+) = \dfrac{0^+}{-\infty} = 0$ , $\quad f(\infty) = \infty$	1(b) Ends of $0 < x < \infty$	Not indeterminate; the sample $f(10^{10}) \approx 4.3 \times 10^{8}$
$f'(x) = \dfrac{\ln x - 1}{(\ln x)^2} = 0 \iff x = e$ , $\quad f(e) = \dfrac{e}{\ln e} = e$	2 Critical points	Quotient rule; a fraction is zero where its numerator is
$\operatorname{sign} f'(x) = \operatorname{sign}(\ln x - 1)$ : $\ -,\ -,\ +$ on $(0,1)$ , $(1,e)$ , $(e,\infty)$	3 Sign of $f'$	$(\ln x)^2 > 0$ ; a double check, not a starting point

Key Insight: Plot the discontinuities and the ends first, then the critical points. With only one critical point, $(e, e)$ , the graph has only one way to turn, and step 3 merely confirms it.

IIVisualization 1 — The Scheme, Step by Step

Plotting alone fixes four features of $x/\ln x$ ; the single critical point $(e, e)$ then fixes the shape.

Step of the scheme 1(a)

IIIVisualization 2 — Numbers at the Trouble Spots

Sizes and signs of numerator and $\ln x$ decide each one-sided limit and each end, with no derivative.

Approach k 4

IVVisualization 3 — Step 3 Is the Double Check

The sign of $f'$ must match the sketch on every interval; a slip in the derivative shows up as a mismatch.

Derivative used x 2.00

VQuiz Questions

Problem 1 · Step 2 for x / ln x

Given: $f(x) = \dfrac{x}{\ln x}$ for $x > 0$ , $x \neq 1$ — find its critical point $(x, f(x))$ .

✅ Correct!

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

vanishes only where

\ln x = 1

, and

f(e) = \dfrac{e}{\ln e} = e

❌ $x = 1$ is not a critical point. There the denominator

\ln x

is zero, so

f

is undefined:

x = 1

is the vertical asymptote from step 1.

❌ The $x$ -value is right; the height is not. The critical value is

f(e) = \dfrac{e}{\ln e} = \dfrac{e}{1} = e

, not

\ln e

❌ A sign slip in the quotient rule. The numerator is

1 \cdot \ln x - x \cdot \tfrac{1}{x} = \ln x - 1

; adding the second term gives

\ln x + 1 = 0

, i.e.

x = \tfrac{1}{e}

❌ Not quite. Differentiate with the quotient rule, set the numerator to zero, then evaluate

f

there.

Show solution

Quotient rule:

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

A fraction is zero where its numerator is zero: $\ln x - 1 = 0 \Rightarrow \ln x = 1 \Rightarrow x = e$ . This is the only critical point.

f(e) = \frac{e}{\ln e} = \frac{e}{1} = e.

The critical point is $(e, e)$ : both coordinates equal $e$ .

Problem 2 · Where the Graph Falls

Given: $f(x) = \dfrac{x}{\ln x}$ with $f'(x) = \dfrac{\ln x - 1}{(\ln x)^2}$ — on which intervals is $f$ decreasing?

✅ Correct! The numerator

\ln x - 1

is negative on both

(0, 1)

and

(1, e)

, and

x = 1

is left out because

f

is undefined there.

❌ $x = 1$ splits the interval.

f

is undefined at

1

and jumps from

-\infty

+\infty

there, so it is not decreasing on all of

(0, e)

: just to the right of

1

the values are far larger than just to the left.

❌ $(0, 1)$ is missing. There

\ln x < 0

, so

\ln x - 1

is even more negative; negative over positive is negative.

❌ That is where $f$ increases. For

x > e

\ln x > 1

, so the numerator is positive and

f' > 0

❌ Not quite. The denominator

(\ln x)^2

is positive, so the sign of

f'

is the sign of

\ln x - 1

Show solution

Split $(0, \infty)$ at the discontinuity $x = 1$ and the critical point $x = e$ . Since $(\ln x)^2 > 0$ , $\operatorname{sign} f' = \operatorname{sign}(\ln x - 1)$ .

$0 < x < 1$ : $\ln x < 0$ , so $\ln x - 1 < 0$ — decreasing.
$1 < x < e$ : $0 < \ln x < 1$ , so $\ln x - 1 < 0$ — decreasing.
$x > e$ : $\ln x > 1$ , so $\ln x - 1 > 0$ — increasing.

So $f$ decreases on $(0, 1)$ and on $(1, e)$ separately, matching the sketch: down from $0$ toward $-\infty$ , then down from $+\infty$ to $(e, e)$ .

Problem 3 · The Scheme on x³ / ln x

Given: $g(x) = \dfrac{x^3}{\ln x}$ for $x > 0$ , $x \neq 1$ — find its critical point and critical value.

Where is the critical point?

What is the critical value?

✅ Correct!

g'(x) = \dfrac{x^2(3\ln x - 1)}{(\ln x)^2}

vanishes at

\ln x = \tfrac{1}{3}

, and

g(e^{1/3}) = \dfrac{e}{1/3} = 3e

❌ Check the critical point. The numerator is

3x^2\ln x - x^2 = x^2(3\ln x - 1)

;

x = 0

is outside the domain, so only

3\ln x = 1

counts, and that gives

\ln x = \tfrac{1}{3}

❌ Check the critical value.

(e^{1/3})^3 = e

and

\ln e^{1/3} = \tfrac{1}{3}

; dividing by

\tfrac{1}{3}

triples.

Show solution

Step 1 (plot): $\ln x = 0$ at $x = 1$ , with $g(1^-) = \tfrac{1}{0^-} = -\infty$ and $g(1^+) = \tfrac{1}{0^+} = +\infty$ ; at the ends, $g(0^+) = \tfrac{0^+}{-\infty} = 0$ and $g(\infty) = \infty$ — the same four features as $x/\ln x$ .

Step 2 (critical points):

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

Since $x > 0$ , $g'(x) = 0 \iff 3\ln x - 1 = 0 \iff \ln x = \tfrac{1}{3} \iff x = e^{1/3}$ .

g(e^{1/3}) = \frac{(e^{1/3})^3}{\ln e^{1/3}} = \frac{e}{1/3} = 3e.

One critical point, $(e^{1/3}, 3e)$ , so the right branch comes down from $+\infty$ , turns at $(e^{1/3}, 3e)$ and climbs back to $\infty$ , exactly as for $x/\ln x$ .

Problem 4 · Catching a Slip at Step 3

Given: A student sketching $f(x) = \dfrac{x}{\ln x}$ writes $f'(x) = \dfrac{1 - \ln x}{(\ln x)^2}$ (quotient rule with the terms reversed), after correctly finding the critical point $(e, e)$ and $f(\infty) = \infty$ . What does step 3 with this formula report on $(e, \infty)$ ?

✅ Correct! For

x > e

1 - \ln x < 0

, so the formula says "decreasing" — but the graph must rise from height

e

\infty

. The contradiction exposes the slip.

❌ Evaluate the student's numerator. For

x > e

\ln x > 1

, so

1 - \ln x

is negative, not positive.

❌ Wrong feature. The asymptote concerns

x

near

1

; on

(e, \infty)

the conflict is with the right end

f(\infty) = \infty

and the height

f(e) = e

❌ The reversed formula has the same zero.

1 - \ln x

and

\ln x - 1

both vanish at

x = e

, so step 2 cannot see the slip; only the signs in step 3 can.

❌ Not quite. Find the sign of

1 - \ln x

for

x > e

and compare with what steps 1 and 2 force.

Show solution

The denominator $(\ln x)^2$ is positive, so the student's formula has the sign of $1 - \ln x$ . For $x > e$ , $\ln x > 1$ , hence $1 - \ln x < 0$ : the formula says $f$ is decreasing on $(e, \infty)$ .

But steps 1 and 2 give $f(e) = e$ and $f(\infty) = \infty$ , and there is no other critical point, so $f$ must increase from $e$ onward. The signs cannot all be true — this is the contradiction step 3 is designed to produce.

The same formula also gives $f' > 0$ on $(0, 1)$ , where the sketch falls from $0$ toward $-\infty$ . Redoing the quotient rule, $\dfrac{(x)' \ln x - x (\ln x)'}{(\ln x)^2} = \dfrac{\ln x - 1}{(\ln x)^2}$ , restores the signs $-,\ -,\ +$ .

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