Chapter 6 · Application of Derivatives

Miscellaneous · Q1: $f(x)=\frac{\log x}{x}$ has its maximum at $x=e$

Miscellaneous ExerciseMaxima and minima · Miscellaneous Exercise

The question

Show that the function given by $f(x) = \dfrac{\log x}{x}$ has maximum at $x = e$ ⁠.

Exam answer

Here $f(x) = \frac{\log x}{x}$ ⁠, $x > 0$ ⁠. By the quotient rule,

f'(x) = \frac{\frac1x\cdot x - \log x}{x^2} = \frac{1 - \log x}{x^2}

$f'(x) = 0 \Rightarrow \log x = 1 \Rightarrow x = e$ ⁠.

For $0 < x < e$ ⁠, $\log x < 1$ ⁠, so $f'(x) > 0$ ⁠; for $x > e$ ⁠, $\log x > 1$ ⁠, so $f'(x) < 0$ ⁠. So $f$ increases up to $x = e$ and decreases after it.

$\therefore\ f$ has its maximum at $x = e$ ⁠, and the maximum value is $f(e) = \dfrac1e$ ⁠. Hence proved.

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

The one idea this problem is built on

If a function rises all the way up to a point and falls all the way after it, then its value at that point is the largest value it ever takes. So we find where $f'$ changes from positive to negative, and we check the sign of $f'$ on the whole domain, not just near the point. That turns "a local maximum" into "the maximum".

Given, to show, and the plan

Given: $f(x) = \dfrac{\log x}{x}$ ⁠, where, as in Chapter 5, $\log x$ means the natural logarithm (base $e$ ⁠).
To show: $f$ has its maximum value at $x = e$ ⁠.
Plan: (1) Find the domain. (2) Differentiate with the quotient rule. (3) Solve $f'(x) = 0$ ⁠. (4) Find the sign of $f'$ on each side. (5) Use Theorem 1 to show $f(e) > f(x)$ for every other $x$ in the domain. (6) Confirm with the second derivative test. (7) Compute the maximum value.

Basics you need

1The natural logarithm

$\log x$ is defined only for $x > 0$ ⁠; $\log 1 = 0$ ⁠, $\log e = 1$ ⁠; and $\log$ is strictly increasing, so $\log x < 1 \iff x < e$ ⁠. Its derivative is $\frac{d}{dx}\log x = \frac{1}{x}$ (Theorem 5 (2), Section 5.4).

2Quotient rule

$\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}$ wherever $v \ne 0$ (Section 5.3, rule (3)).

3Theorem 1, Section 6.3

If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ ⁠, then $f' > 0$ on $(a, b)$ makes $f$ increasing on $[a, b]$ ⁠, and $f' < 0$ makes it decreasing. The Mean Value step in the book's proof, $f(x_2) - f(x_1) = f'(c)(x_2 - x_1)$ ⁠, even gives strict inequalities. Lesson: increasing and decreasing functions

4Maximum value and the second derivative test

Definition 3, Section 6.4: $f$ has its maximum value on I at $c$ when $f(c) \ge f(x)$ for every $x$ in I. Theorem 4 (second derivative test): $f'(c) = 0$ and $f''(c) < 0$ make $c$ a point of local maxima. Lesson: derivative tests

5Sign of a product or quotient

A quotient with a positive denominator has the sign of its numerator. Basics: sign of a product

Solution

The domain is $x > 0$

$\log x$ exists only for $x > 0$ ⁠, and then the denominator $x$ is not zero either. So the domain of $f$ is the interval
$(0, \infty).$
On it, $f$ is a quotient of two differentiable functions with a non-zero denominator, so $f$ is continuous and differentiable at every point.

"Maximum" always means "maximum on some set". Here the only sensible set is the whole domain, so we must know what it is. Negative $x$ and $x = 0$ are simply not available.

Natural logarithm: defined for positive numbers only (Basic 1).
Differentiate with the quotient rule

Take $u = \log x$ and $v = x$ ⁠, so $u' = \frac1x$ and $v' = 1$ ⁠:
$\begin{aligned} f'(x) &= \frac{u'v - uv'}{v^2} \\ &= \frac{\frac1x \cdot x - \log x \cdot 1}{x^2} \\ &= \frac{1 - \log x}{x^2}. \end{aligned}$
$\frac1x \cdot x = 1$ for every $x \ne 0$ ⁠, which is what makes the numerator so simple. The quotient rule is allowed because $v = x \ne 0$ on the domain.

Quotient rule (Basic 2) and $\frac{d}{dx}\log x = \frac1x$ (Basic 1).
The only critical point is $x = e$
$\begin{aligned} f'(x) = 0 &\iff 1 - \log x = 0 \\ &\iff \log x = 1 \\ &\iff x = e. \end{aligned}$
Since $f$ is differentiable at every point of its domain, there are no critical points of the other kind, so $x = e$ is the only critical point.

A fraction is zero exactly when its numerator is zero (the denominator $x^2$ is positive here). $\log x = 1$ has the single solution $x = e$ ⁠, because $\log e = 1$ and $\log$ is one-one. By Theorem 2, Section 6.4, a maximum inside the domain can only occur at a critical point, so $x = e$ is the only candidate.

Critical point: $f'(c) = 0$ or $f$ not differentiable at $c$ (Note after Theorem 2, Section 6.4). Lesson: maxima and minima
Sign of $f'$ ⁠: positive before $e$ ⁠, negative after

The denominator $x^2$ is positive, so $f'(x)$ has the sign of $1 - \log x$ ⁠:
- for $0 < x < e$ ⁠: $\log x < \log e = 1$ ⁠, so $1 - \log x > 0$ and $f'(x) > 0$ ⁠;
- for $x > e$ ⁠: $\log x > 1$ ⁠, so $f'(x) < 0$ ⁠.
Test points: $f'(1) = \frac{1 - 0}{1} = 1 > 0$ and $f'(e^2) = \frac{1 - 2}{e^4} = -\frac{1}{e^4} < 0$ ⁠.

Sign chart on the domain $(0, \infty)$ ⁠, not to scale (the open circle at 0 marks that 0 is not in the domain): $f'$ changes from $+$ to $-$ at $x = e$ ⁠.

Comparing $\log x$ with 1 is the same as comparing $x$ with $e$ ⁠, because $\log$ is strictly increasing. So the sign question reduces to "is $x$ less than or greater than $e$ ⁠?".

Positive denominator keeps the sign of the numerator (Basic 5). Basics: sign of a product
$f(e)$ beats every other value on the whole domain

Left of $e$ ⁠. Take any $x$ with $0 < x < e$ ⁠. $f$ is continuous on $[x, e]$ and $f' > 0$ on $(x, e)$ ⁠, so by Theorem 1 (with the strict inequality from its proof)
$f(x) < f(e).$
Right of $e$ ⁠. Take any $x > e$ ⁠. $f$ is continuous on $[e, x]$ and $f' < 0$ on $(e, x)$ ⁠, so $f$ is decreasing there and
$f(e) > f(x).$
Together: $f(e) > f(x)$ for every $x \ne e$ in $(0, \infty)$ ⁠. By Definition 3, Section 6.4, $f$ has its maximum value on its domain at $x = e$ , and at no other point.

The graph of $y = \frac{\log x}{x}$ (horizontal and vertical scales differ). It climbs steeply, crosses the $x$ -axis at $x = 1$ ⁠, peaks at $\left(e, \frac1e\right)$ and then sinks slowly towards 0 without reaching it.

Knowing only that $x = e$ is a local maximum would say that $f(e)$ beats the values near $e$ ⁠. The question says "maximum", and the sign of $f'$ on the whole domain is what lets us compare $f(e)$ with values far away, such as $f(0.01)$ or $f(1000)$ ⁠.

Theorem 1, Section 6.3 (Basic 3); Definition 3, Section 6.4 (Basic 4). Lesson: absolute extrema
Cross-check with the second derivative test

Differentiate $f'(x) = \frac{1 - \log x}{x^2}$ by the quotient rule, with $u = 1 - \log x$ ⁠, $u' = -\frac1x$ ⁠, $v = x^2$ ⁠, $v' = 2x$ ⁠:
$\begin{aligned} f''(x) &= \frac{-\frac1x \cdot x^2 - (1 - \log x)\cdot 2x}{x^4} \\ &= \frac{-x - 2x + 2x\log x}{x^4} \\ &= \frac{2\log x - 3}{x^3}. \end{aligned}$
At $x = e$ ⁠: $f''(e) = \frac{2 - 3}{e^3} = -\frac{1}{e^3} < 0$ ⁠. With $f'(e) = 0$ ⁠, Theorem 4 (i) says $x = e$ is a point of local maxima, which agrees with Step 5.

This test is quick but local: it only describes $f$ near $e$ ⁠. It confirms Step 5 but could not replace it. In the last line we divided top and bottom by $x$ ⁠, which is allowed because $x > 0$ ⁠.

Second derivative test, Theorem 4, Section 6.4 (Basic 4). Lesson: derivative tests
The maximum value is $\frac1e$
$f(e) = \frac{\log e}{e} = \frac{1}{e} \approx 0.3679.$
The question only asks where the maximum is, but stating the value completes the picture and lets us test the claim with numbers.

Result

On its domain $(0, \infty)$ ⁠, $f'(x) = \dfrac{1 - \log x}{x^2}$ is positive for $x < e$ and negative for $x > e$ ⁠. So $f$ increases up to $x = e$ and decreases after it, and $f(e) > f(x)$ for every $x \ne e$ ⁠: $f$ has its maximum at $x = e$ , with maximum value $f(e) = \dfrac1e$ ⁠. (The answer key has no entry for this "show that" question.)

Check it with numbers

$f(e) = \frac1e \approx 0.36788$ ⁠. Nearby and far away, $f$ is smaller: $f(2) = \frac{\log 2}{2} \approx 0.34657$ ⁠, $f(3) = \frac{\log 3}{3} \approx 0.36620$ ⁠, $f(10) \approx 0.23026$ ⁠, $f(0.5) \approx -1.38629$ ⁠. A nice by-product: $f(\pi) < f(e)$ means $\frac{\log \pi}{\pi} < \frac{1}{e}$ ⁠, that is $e\log\pi < \pi$ ⁠, so $\pi^e < e^\pi$ ⁠. Indeed $\pi^e \approx 22.459$ and $e^\pi \approx 23.141$ ⁠.

Common mistakes

Stopping at $f'(e) = 0$ ⁠. A zero derivative only makes $e$ a candidate; $x^3$ has $f'(0) = 0$ and no maximum there (Remark after Theorem 2).
Using only the second derivative test and calling the result "the maximum". That test is local; the sign of $f'$ on the whole domain is what proves the value is the largest.
Forgetting the domain and talking about $x \le 0$ ⁠, where $\log x$ does not exist.
Dropping the $-$ in the quotient rule, which gives $\frac{1 + \log x}{x^2}$ and the wrong critical point $x = \frac1e$ ⁠.

What to take away

" $f'$ positive up to $c$ ⁠, negative after $c$ ⁠, on the whole domain" proves that $f(c)$ is the greatest value, not just a local one.
When the numerator of $f'$ is $1 - \log x$ ⁠, compare $x$ with $e$ ⁠: $\log$ is increasing and $\log e = 1$ ⁠.
The maximum of $\frac{\log x}{x}$ at $e$ explains why $e^\pi > \pi^e$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
$\log x$ means $\log_e x$ ⁠; defined for $x > 0$ ⁠; strictly increasing; $\log e = 1$	Steps 1, 3, 4	Section 5.4
$\frac{d}{dx}\log x = \frac1x$	Steps 2, 6	Theorem 5 (2), Section 5.4
Quotient rule	Steps 2, 6	Section 5.3, rule (3)
Critical point; extrema occur only at critical points	Step 3	Theorem 2 and the Note after it, Section 6.4
$f' > 0$ (or $< 0$ ⁠) on $(a, b)$ gives increasing (decreasing) on $[a, b]$ ⁠, strictly	Step 5	Theorem 1 and its proof, Section 6.3
Maximum value: $f(c) \ge f(x)$ for all $x$ in I	Step 5	Definition 3 (a), Section 6.4 (the inequality sign in this reprint looks like " $>$ "; it must be " $\ge$ ", since $x = c$ is allowed)
Second derivative test	Step 6	Theorem 4 (i), Section 6.4
All decimal values	Step 7, Check	Recomputed in the site's verification script

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Miscellaneous · Q1: f(x)=log⁡xxf(x)=\frac{\log x}{x}f(x)=xlogx​ has its maximum at x=ex=ex=e

Basics you need

Solution

The domain is x>0x > 0x>0

Differentiate with the quotient rule

The only critical point is x=ex = ex=e

Sign of f′f'f′⁠: positive before eee⁠, negative after

f(e)f(e)f(e) beats every other value on the whole domain

Cross-check with the second derivative test

The maximum value is 1e\frac1ee1​

Every fact used, and where it comes from

Miscellaneous · Q1: $f(x)=\frac{\log x}{x}$ has its maximum at $x=e$

The domain is $x > 0$

The only critical point is $x = e$

Sign of $f'$ ⁠: positive before $e$ ⁠, negative after

$f(e)$ beats every other value on the whole domain

The maximum value is $\frac1e$