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

Concept: Derivative Sign Analysis — to find a maximum, we check where $f'(x) = 0$ and verify the sign change of $f'(x)$ from positive to negative.

Step 1: Differentiate $f(x) = \frac{\log x}{x}$ using the quotient rule:

Step 2: Set $f'(x) = 0$. The denominator $x^2 > 0$ for $x > 0$, so:

Step 3: Check the sign of $f'(x)$ around $x = e$. For $x < e$ (say $x = 1$), $\log x < 1$, so $1 - \log x > 0$ → $f'(x) > 0$. For $x > e$ (say $x = 3$), $\log x > 1$, so $1 - \log x < 0$ → $f'(x) < 0$. Since $f'(x)$ changes from positive to negative at $x = e$, the function has a local maximum there.

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:

The domain is $x > 0$ because $\log x$ is defined only for positive $x$.

2. Set the derivative to zero.

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 $x < e$ (say $x = 2$): $\log 2 \approx 0.693 < 1$, so $1 - \log x > 0$. Hence $f'(x) > 0$ — the function is increasing.
   • For $x > e$ (say $x = 3$): $\log 3 \approx 1.099 > 1$, so $1 - \log x < 0$. Hence $f'(x) < 0$ — the function is decreasing.

   So $f'(x)$ changes from positive to negative at $x = e$.

4. Conclude the nature of the critical point. Since $f'(x) > 0$ for $x < e$ and $f'(x) < 0$ for $x > e$, the function increases up to $x = e$ and then decreases after. Therefore, $x = e$ is a point of local maximum.

Method: Locating and Classifying a Critical Point with the First-Derivative Sign Test

This method proves that a specific point is a maximum (or minimum) of a function by tracking how the sign of the derivative changes around it — rather than relying only on the second derivative.

Steps

Step 1: Differentiate the function using the appropriate rule

For a quotient like $f(x)=\dfrac{\log x}{x}$, apply the quotient rule carefully:

Step 2: Set the derivative to zero and solve within the function's domain

Since the denominator $x^2$ is never zero for $x$ in the domain, focus on the numerator:

Step 3: Determine the sign of $f'(x)$ just below and just above the critical point

Because the denominator $x^2$ is always positive on the domain, the sign of $f'(x)$ is decided entirely by the numerator $1-\log x$. Pick a test value less than the critical point and one greater, and check whether the numerator is positive or negative at each.

Step 4: Read off the conclusion from the sign change

• Sign changes from positive to negative as $x$ increases through the critical point $\implies$ local maximum there.
• Sign changes from negative to positive $\implies$ local minimum.
• No sign change $\implies$ not an extremum at all (just a flat inflection).

Step 5: State the conclusion explicitly, matching what was asked

If the question asks you to show a maximum exists at a given point, explicitly state both the sign of $f'$ on each side and the resulting conclusion — a bare "$f'(x)=0$ at $x=e$" on its own does not prove a maximum.

Common Mistakes

Mistake 1: A quotient-rule sign error

Writing $f'(x)=\dfrac{1+\log x}{x^2}$ instead of the correct $\dfrac{1-\log x}{x^2}$ (dropping the minus sign that comes from differentiating $\log x$ in the numerator) is a very common slip, and it shifts the critical point to the wrong location entirely.

Mistake 2: Concluding "maximum" just because $f'(x)=0$

A zero derivative only identifies a candidate critical point — it does not by itself prove a maximum. Without explicitly checking that $f'(x)$ changes from positive to negative around $x=e$ (or checking $f''(e)<0$), the claim "$f$ has a maximum at $x=e$" is asserted, not shown, which is exactly what the question asks you to demonstrate.

Mistake 3: Confusing $\log x$ with base-10 logarithm

In this NCERT/CBSE calculus context, $\log x$ denotes the natural logarithm ($\ln x$), which is why the critical point lands exactly on $x=e$. Treating it as $\log_{10}x$ would shift the critical point to $x = 10^{\log_{10}e}$ — a different, and here incorrect, value.