# Thread: ABS Max and Min

1. ## ABS Max and Min

Find the absolute maximum and absolute minimum of the function $f(x)=ln(3x)/x$ on the interval [0.1, 5]

2. Originally Posted by Shivian
Find the absolute maximum and absolute minimum of the function $f(x)=ln(3x)/x$ on the interval [0.1, 5]

Derivate and put f'(x)=0 to find candidates to be the maximum.

Tonio

3. Is this correct?

$x=(e^9)-3$

4. Originally Posted by tonio
Derivate and put f'(x)=0 to find candidates to be the maximum.

Tonio
He will also need to check the boundary points.

5. Originally Posted by Bruno J.
He will also need to check the boundary points.

Of course. Thanx for the clarification.

Tonio

6. Originally Posted by Shivian
Is this correct?

$x=(e^9)-3$

I don't think so. What's the derivative of your function, anyway?

Tonio

7. Here is all I have so far.

Interval: [0.1, 5]
$f(x)=ln(3x)/x$
$f1(x)=(1-ln(3x))/(x^2)$
$x=e/3$

$f(.1)=-12.0397$
$f(5)=0.5416$
$f(e/3)=1.1036$

Max is $f(e/3)=1.1036$
Min is $f(.1)=-12.0397$

How close am I?

8. Originally Posted by Shivian
Here is all I have so far.

Interval: [0.1, 5]
$f(x)=ln(3x)/x$
$f1(x)=(1-ln(3x))/(x^2)$
$x=e/3$

$f(.1)=-12.0397$
$f(5)=0.5416$
$f(e/3)=1.1036$

Max is $f(e/3)=1.1036$
Min is $f(.1)=-12.0397$

How close am I?
So close that you already are there: I see no mistake in the above at all.
Way to go!

Tonio