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

> [!TLDR] > The function $f(x) = \frac{\log x}{x}$ attains its maximum at $x = e$ because its derivative changes sign from positive to negative at that point — the classic test for a local maximum. We want to show that $f(x) = \frac{\log x}{x}$ has a maximum at $x = e$. The natural approach is to use derivative sign analysis: find where $f'(x) = 0$, then check whether the derivative changes from positive to negative at that point. That change tells us the function stops increasing and starts decreasing — the hallmark of a local maximum. Why does this work? A function's derivative tells us its slope. If $f'(x) > 0$, the function is rising; if $f'(x) < 0$, it's falling. At a maximum, the function transitions from rising to falling, so the derivative must go from positive to zero to negative. The point where $f'(x) = 0$ is a candidate; the sign change confirms it. Let's work through it step by step. 1. **Find the derivative.** $f(x) = \frac{\log x}{x}$. Here $\log x$ is the natural logarithm (base $e$). Use the quotient rule: $$f'(x) = \frac{(1/x) \cdot x - \log x \cdot 1}{x^2} = \frac{1 - \log x}{x^2}.$$ The domain is $x > 0$ because $\log x$ is defined only for positive $x$. 2. **Set the derivative to zero.** $$f'(x) = 0 \implies \frac{1 - \log x}{x^2} = 0 \implies 1 - \log x = 0 \implies \log x = 1.$$ Since $\log e = 1$, we get $x = e$. So $x = e$ is the only critical point in the domain. 3. **Analyze the sign of $f'(x)$ around $x = e$.** The denominator $x^2$ is always positive for $x > 0$, so the sign of $f'(x)$ depends entirely on the numerator $1 - \log x$. - For $x 0$. Hence $f'(x) > 0$ — the function is increasing. - For $x > e$ (say $x = 3$): $\log 3 \approx 1.099 > 1$, so $1 - \log x [!WARNING] > A common mistake is to forget that $\log

Show that the function given by f(x) = (log x)/(x) has maximum at x = e.

Miscellaneous Exercise · Q1

f(x) = \frac{\log x}{x}

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Concept understanding — Derivative Sign Analysis

Derivative Sign Analysis: What the Slope Tells You

Imagine walking along a hilly road — sometimes uphill, sometimes downhill, occasionally flat. The derivative at any point is simply the slope of the road under your feet at that instant.

Derivative sign analysis figures out where a function is increasing, where it is decreasing, and where it has flat spots (critical points) — all from the sign of its derivative.

The Intuition First

If $f'(x)$ is positive, the function is increasing — the graph rises as you move right. If $f'(x)$ is negative, it is decreasing. If $f'(x) = 0$ , there is a horizontal tangent — a potential peak, valley, or flat inflection.

The key: a single point tells you little; you look at intervals. If $f'(x) > 0$ for all $x$ in $(a, b)$ , the function is strictly increasing on that whole interval. Same logic for negative.

Note

The analysis is local — it describes behaviour on intervals, not isolated points. A zero derivative at a single point doesn't guarantee a max or min; check the sign change across that point.

The Precise Statement

Let $f$ be differentiable on an open interval $I$ . Then:

If $f'(x) > 0$ for all $x$ in $I$ , then $f$ is strictly increasing on $I$ .
If $f'(x) < 0$ for all $x$ in $I$ , then $f$ is strictly decreasing on $I$ .
If $f'(x) = 0$ for all $x$ in $I$ , then $f$ is constant on $I$ .

Points where $f'(x) = 0$ (or where $f'$ does not exist) are critical points — the candidates for local maxima and minima.

\text{If } f'(x) > 0 \text{ on } (a,b) \implies f \text{ increasing on } (a,b)

How to Perform It (Step-by-Step)

Find the derivative $f'(x)$ .
Find critical points: solve $f'(x) = 0$ and check where $f'(x)$ is undefined (but $f$ is defined).
Plot these on a number line — they split the domain into intervals.
Pick a test point inside each interval and evaluate $f'$ ; only the sign matters.
Record the sign in each interval and interpret: + means increasing, – means decreasing.

A Concrete Example

Take $f(x) = x^3 - 3x$ .

Step 1: $f'(x) = 3x^2 - 3 = 3(x-1)(x+1)$ .

Step 2: Critical points: $x = -1$ and $x = 1$ .

Step 3: Intervals: $(-\infty, -1)$ , $(-1, 1)$ , $(1, \infty)$ .

Step 4: Test points:

$x = -2$ : $f'(-2) = 3(4-1) = 9 > 0$ .
$x = 0$ : $f'(0) = -3 < 0$ .
$x = 2$ : $f'(2) = 9 > 0$ .

Step 5: So $f$ increases on $(-\infty, -1)$ , decreases on $(-1, 1)$ , increases on $(1, \infty)$ . Thus $x = -1$ is a local maximum (sign changes + to –), and $x = 1$ is a local minimum (– to +).

Watch out

A common mistake: assuming $f'(x) = 0$ automatically means a max or min. Consider $f(x) = x^3$ at $x=0$ : the derivative is zero, but the function increases on both sides (no sign change). That's a saddle point, not an extremum.

Why This Matters for Exams

Derivative sign analysis is the backbone of finding intervals of increase/decrease, locating local maxima/minima (First Derivative Test), sketching graphs, and solving optimization problems.

Tip

Factor the derivative completely. Then the sign of $f'(x)$ follows from the signs of its factors — you can often skip plugging in numbers by reasoning about factor signs on each interval.

Sign analysis of the first derivative to locate increasing/decreasing intervals and critical points is one of the most exam-relevant procedures in the NCERT Class 12 Application of Derivatives chapter, appearing in CBSE boards, JEE Main and as a warm-up for the First Derivative Test. Students searching 'derivative sign chart method' or 'increasing decreasing intervals using derivatives class 12 examples' will find this factor-and-test-point routine is exactly the standard step-by-step technique.

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Q.Show that the function given by f(x)=log⁡xxf(x) = \frac{\log x}{x}f(x)=xlogx​ has maximum at x=ex = ex=e.