1. ## Absolute Minimum

Find the -coordinate of the absolute minimum for the function

I found the derivative and set it equal to 0 and solved for x and got that x=e^(11/3)

Isn't that where the absolute minimum occurs?

my homework says it is wrong.

could i get some help please.

thank you

2. Thats the same thing I came up with

3. You have probably found the LOCAL minimum. The absolute minimum doesn't exist because as x approaches 1, y approaches neg infinity (ie just keeps getting small). Have a look at the graph.

4. i did look at the graph and as x increases y decreases, so there is never a minimum

but my webwork says its a number because i tried entering DNE and -infinity but neither of those work.

5. Originally Posted by Debsta
You have probably found the LOCAL minimum. The absolute minimum doesn't exist because as x approaches 1, y approaches neg infinity (ie just keeps getting small). Have a look at the graph.
Actually if you look at the graph past $\displaystyle e^\frac{11}{3}$ it starts increasing again

6. Originally Posted by 11rdc11
Actually if you look at the graph past $\displaystyle e^\frac{11}{3}$ it starts increasing again

I must be graphing it wrong cause i see a decreasing curve (looks like the left half of a U) that continues to decrease as x goes to infinity.

here is how i enter it into my calculator

(8-3ln(x))/(x)

7. i just entered e^11/3 in again and it worked? weird lol

thanks everyone

8. Zoom in. The LOCAL min occurs when x = e^(11/3) ie approx 39.13.
The question though is asking for the ABSOLUTE min.