Question 67 of 82intermediate🔧 ApplyLong Answer5 marks

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

Correct Answer

Given function is:

f(x)=log⁡xxf(x) = \frac{\log x}{x}

Step 1: Find the first derivative

Differentiating f(x)f(x) with respect to xx using quotient rule:

f′(x)=x⋅1x−log⁡x⋅1x2f'(x) = \frac{x \cdot \frac{1}{x} - \log x \cdot 1}{x^2} f′(x)=1−log⁡xx2f'(x) = \frac{1 - \log x}{x^2}

Step 2: Find critical points

For maxima or minima, f′(x)=0f'(x) = 0:

1−log⁡xx2=0\frac{1 - \log x}{x^2} = 0 1−log⁡x=01 - \log x = 0 log⁡x=1\log x = 1 x=ex = e

Step 3: Apply second derivative test

Differentiating f′(x)f'(x) again:

f′′(x)=x2⋅(−1x)−(1−log⁡x)⋅2xx4f''(x) = \frac{x^2 \cdot \left(-\frac{1}{x}\right) - (1 - \log x) \cdot 2x}{x^4} f′′(x)=−x−2x+2xlog⁡xx4f''(x) = \frac{-x - 2x + 2x \log x}{x^4} f′′(x)=−3+2log⁡xx3f''(x) = \frac{-3 + 2 \log x}{x^3}

At x=ex = e:

f′′(e)=−3+2log⁡ee3f''(e) = \frac{-3 + 2 \log e}{e^3} f′′(e)=−3+2e3=−1e3<0f''(e) = \frac{-3 + 2}{e^3} = \frac{-1}{e^3} < 0

Since f′′(e)<0f''(e) < 0, by the second derivative test, x=ex = e is a point of local maximum for the function f(x)=log⁡xxf(x) = \frac{\log x}{x}.

Exercise: Miscellaneous Exercise on Chapter 6 | Q: 1 | (Page 37)
For More Understanding

Explanation

This question directly appears in the Miscellaneous Exercise on Chapter 6 (Question 1). The solution follows the standard procedure for finding maxima using derivatives as demonstrated in the profit maximization example in the context.

The methodology involves: (1) finding the first derivative, (2) setting it to zero to find critical points, and (3) applying the second derivative test where f′′(x)<0f''(x) < 0 indicates a maximum.

This is consistent with the textbook's approach shown in the profit function example where P′′(240)<0P''(240) < 0 confirmed a point of maxima.

Solution Steps

  1. Step 1: Find f′(x)f'(x) using quotient rule: f′(x)=1−log⁡xx2f'(x) = \frac{1 - \log x}{x^2}

  2. Step 2: Set f′(x)=0f'(x) = 0 to find critical point: x=ex = e

  3. Step 3: Find f′′(x)f''(x) and evaluate at x=ex = e: f′′(e)=−1e3<0f''(e) = \frac{-1}{e^3} < 0

  4. Step 4: Conclude that since f′′(e)<0f''(e) < 0, x=ex = e is a point of maximum

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