Thread: AP calculus exam question on Absolute max

1. $\text{22. Let f be the function defined by$f(x)=\dfrac{\ln x}{x}$What is the absolute maximum value of f ? }$
$$(A)\, 1\quad (B)\, \dfrac{1}{e} (C)\, 0 \quad (D) -e \quad (E) f\textit{ does not have an absolute maximum value}.$$

I only guessed this by graphing it and it appears to $\dfrac{1}{e}$ which is (B)

2.

3. Originally Posted by karush
$\text{22. Let f be the function defined by$f(x)=\dfrac{\ln x}{x}$What is the absolute maximum value of f ? }$
$$(A)\, 1\quad (B)\, \dfrac{1}{e} (C)\, 0 \quad (D) -e \quad (E) f\textit{ does not have an absolute maximum value}.$$

I only guessed this by graphing it and it appears to $\dfrac{1}{e}$ which is (B)
Why would graphing be a guess? It's a valid Mathematical tool!

You could do this by taking the derivative of f(x) and finding the critical points, etc. But if you have this question on an exam the simplest (and probably fastest) way is to take a look at each answer and see what you get. D) is out because f(x) takes on positive values, and A), C), and E) are out by looking at the graph. That leaves B).

-Dan

4. $f’(x)=\dfrac{x \cdot \frac{1}{x} - \ln{x} \cdot 1}{x^2} = \dfrac{1-\ln{x}}{x^2}$

$f’(x)=0$ at $x=e$

first derivative test ...

$x < e \implies f’(x) > 0 \implies f(x) \text{ increasing over the interval } (0,e)$

$x > e \implies f’(x) < 0 \implies f(x) \text{ decreasing over the interval } (e, \infty)$

conclusion ... $f(e) = \dfrac{1}{e}$ is an absolute maximum.

second derivative test ...

$f’’(x) = \dfrac{x^2 \cdot \left(-\frac{1}{x} \right) - (1-\ln{x}) \cdot 2x}{x^4} = \dfrac{2\ln{x} - 3}{x^3}$

$f’’(e) = -\dfrac{1}{e^3} < 0 \implies f(e) = \dfrac{1}{e}$ is a maximum.

wow that was a lot of help..

yes the real negative about these assessment tests is how fast you can eliminate possible answers
not so much what math steps are you really need to take

actually Im learning a lot here at MHB
Mahalo