Calculus: Maximum Value of f(x) = \frac{\log_e x}{x}$

The problem asks us to find the value of $a^2f''(a)$, where $a$ is the value of $x$ at which the function $f(x) = \frac{\log_e x}{x}$ (for $x > 0$) achieves its maximum value. We will use calculus, specifically derivatives, to solve this.

Finding the Maximum Value Point a$

To find where the function reaches its maximum, we need to calculate its first derivative, $f'(x)$, and find the critical points by setting $f'(x) = 0$. The function is $f(x) = \frac{\log_e x}{x}$.

Using the quotient rule $\left(\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}\right)$, let $u = \log_e x$ and $v = x$. Then $u' = \frac{1}{x}$ and $v' = 1$.

The first derivative is:

$f'(x) = \frac{(\frac{1}{x})(x) - (\log_e x)(1)}{x^2}$ $f'(x) = \frac{1 - \log_e x}{x^2}$

To find the critical points, we set $f'(x) = 0$:

$ \frac{1 - \log_e x}{x^2} = 0 $

This equation holds true when the numerator is zero (since $x > 0$, $x^2$ is never zero):

$ 1 - \log_e x = 0 $ $ \log_e x = 1 $

Solving for $x$, we find the critical point:

$ x = e $

Thus, the maximum value of the function occurs at $x = a = e$.

Calculating the Second Derivative f''(x)$

Next, we need to find the second derivative, $f''(x)$, to evaluate it at $x=a$. We differentiate $f'(x) = \frac{1 - \log_e x}{x^2}$ using the quotient rule again.

Let $u = 1 - \log_e x$ and $v = x^2$. Then $u' = -\frac{1}{x}$ and $v' = 2x$.

The second derivative is:

$f''(x) = \frac{(-\frac{1}{x})(x^2) - (1 - \log_e x)(2x)}{(x^2)^2}$ $f''(x) = \frac{-x - 2x + 2x\log_e x}{x^4}$

Simplify the expression:

$f''(x) = \frac{-3x + 2x\log_e x}{x^4}$

Factor out $x$ from the numerator and cancel with the denominator:

$f''(x) = \frac{x(-3 + 2\log_e x)}{x^4}$ $f''(x) = \frac{2\log_e x - 3}{x^3}$

Evaluating f''(a)$

We substitute $a = e$ into the second derivative formula:

$f''(e) = \frac{2\log_e e - 3}{e^3}$

Since $\log_e e = 1$:

$f''(e) = \frac{2(1) - 3}{e^3}$ $f''(e) = \frac{2 - 3}{e^3}$ $f''(e) = \frac{-1}{e^3}$

The negative value of $f''(e)$ confirms that $x=e$ is indeed a point of maximum.

Calculating the Final Value a^2f''(a)$

Finally, we compute $a^2f''(a)$ using $a=e$ and $f''(e) = -\frac{1}{e^3}$:

$a^2f''(a) = e^2 \times f''(e)$ $a^2f''(a) = e^2 \times \left(-\frac{1}{e^3}\right)$ $a^2f''(a) = -\frac{e^2}{e^3}$ $a^2f''(a) = -\frac{1}{e}$