Question

The maximum values of \(\frac{\log _{e} x}{x}\) is (1) 1 (2) \(\frac{2}{e}\) (3) \(\mathrm{e}\) (4) \(\frac{1}{e}\)

Short answer

The maximum value is \( \frac{1}{e} \).

Step-by-step solution

Define the function

Let the function be defined as \[ f(x) = \frac{\log_e x}{x} \]

Find the derivative

To find the critical points, first calculate the derivative of the function. Using the quotient rule where \( u = \log_e x \) and \( v = x \):\[ f'(x) = \frac{v \cdot \frac{d}{dx}(u) - u \cdot \frac{d}{dx}(v)}{v^2} = \frac{x \cdot \frac{1}{x} - \log_e x \cdot 1}{x^2} = \frac{1 - \log_e x}{x^2} \]

Find critical points

Set the derivative equal to zero to find the critical points:\[ \frac{1 - \log_e x}{x^2} = 0 \]This implies:\[ 1 - \log_e x = 0 \rightarrow \log_e x = 1 \rightarrow x = e \]

Determine max value

Evaluate the function at the critical point to determine the maximum value:\[ f(e) = \frac{\log_e e}{e} = \frac{1}{e} \]This is the maximum value.

Verify the second derivative

For confirmation, check the second derivative to ensure it is a maximum:Find the second derivative:\[ f''(x) = \frac{d}{dx} \left( \frac{1 - \log_e x}{x^2} \right) \]. Using the quotient rule again,\[ f''(x) = \frac{\frac{d}{dx}(1 - \log_e x) \cdot x^2 - (1 - \log_e x) \cdot \frac{d}{dx}(x^2)}{(x^2)^2} = \frac{-\frac{1}{x} \cdot x^2 - (1 - \log_e x) \cdot 2x}{x^4} \], which simplifies to:\[ f''(x) = \frac{-x - 2x + 2x \log_e x}{x^4} = \frac{-3x + 2x \log_e x}{x^4} = \frac{-3 + 2 \log_e x}{x^3} \]. Evaluating at \( x = e \),\[ f''(e) = \frac{-3 + 2 \cdot 1}{e^3} = \frac{-1}{e^3} \], which is negative, confirming a local maximum at \( x = e \).

Key concepts

Derivative

The derivative is a fundamental concept in calculus. It measures how a function changes as its input changes. Think of it as the slope of the function at any given point.

The derivative is represented as \( f'(x) \) or \( \frac{d}{dx} f(x) \), and it gives the rate of change of the function \( f(x) \). For the function \( f(x) = \frac{\log_e x}{x} \), this tells us how \( \frac{\log_e x}{x} \) changes with respect to \( x \).

To find the derivative, we use different rules depending on the function's form. In this problem, we used the Quotient Rule because the function is a ratio of two expressions.

Critical Points

Finding critical points is an essential step in optimization problems. A critical point occurs where the derivative of the function is zero or undefined.

To find the critical points for \( f(x) = \frac{\log_e x}{x} \), we set \( f'(x) = 0 \). This is because, at critical points, the slope of the function is horizontal. From our solution, we found:
\[ \frac{1 - \log_e x}{x^2} = 0 \]
This simplifies to:\[ 1 - \log_e x = 0 \rightarrow \log_e x = 1 \rightarrow x = e \]. So, the critical point is \( x = e \).

Second Derivative Test

The Second Derivative Test helps to determine if a critical point is a maximum, minimum, or saddle point. After finding the critical points, we take the second derivative of the function and evaluate it at those points.

If \( f''(x) \) is positive, the point is a local minimum. If \( f''(x) \) is negative, the point is a local maximum. For our function, we found:
\[ f''(e) = \frac{-1}{e^3} \]
Since this is negative, it confirms that \( x = e \) is a point of local maximum.

Quotient Rule

The Quotient Rule is used for finding the derivative of a function that is the ratio of two differentiable functions. For a function \( f(x) = \frac{u(x)}{v(x)} \), its derivative is derived as:
\[ f'(x) = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{v(x)^2} \]
Using this, we calculated the derivative of \( f(x) = \frac{\log_e x}{x} \) where \( u(x) = \log_e x \) and \( v(x) = x \), leading to:
\[ f'(x) = \frac{x \cdot \frac{1}{x} - \log_e x \cdot 1}{x^2} = \frac{1 - \log_e x}{x^2} \]. This was the key step in finding the critical points and optimizing the function.