You have the correct derivative. One thing you can do is let
Then you have
Now, solve for u and remember that
Hello, I'm having trouble with using the max and min algorithm (finding the derivative, setting it to 0, and isolating) to find absolute max and min values for log and exponential functions. My teacher mentioned something about taking the ln of both sides when you have e, and taking the e of both sides when you have ln- however, I'm confused as to how to do this. For example, this question:
Determine the absolute max and min values for this function: (e^-x) - (e^-3x), where x is less than or equal to 10, and greater than or equal to 0.
I found the derivative to be -e^-x + 3e^-3x. Now I know I need to set it 0, and solve for x- however I'm thrown off by the e base. How do I solve for x when I have e? And vice versa, if I had ln, how would I solve for x?
I'd greatly appreciate any help, especially as I have to know this for a test tomorrow. Thank you in advance!
You need to brush up on your PreCalculus . . .
Determine the absolute max and min values for:
I found the derivative to be: . . . . . Good!
We have: .
Take the log of both sides: .
. . and we have: . . . . . a critical value.
The second derivative is: .
When . . . . negative
. . Hence, the graph is concave down; we have a maximum point.
We must also examine the endpoints of the interval.