Intervals of increase/decrease with a exponential function

Click For Summary

Homework Help Overview

The discussion revolves around finding the intervals of increase and decrease for the exponential function f(x) = e3x + e-2x. Participants explore the process of determining critical points through the derivative of the function.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to find critical numbers by setting the derivative f'(x) = 0 and expresses confusion over an apparent mistake in their calculations. Other participants question the validity of logarithmic manipulations used in the original approach.

Discussion Status

The discussion includes reflections on mistakes made during the problem-solving process, with participants sharing personal experiences of similar errors. There is an ongoing exploration of the correct application of logarithmic properties and differentiation techniques.

Contextual Notes

Participants express frustration over simple mistakes and mix-ups in formulas, indicating a common challenge in understanding the differentiation process and its implications for solving the problem.

endeavor
Messages
174
Reaction score
0
find the intervals of increase and decrease of f(x) = e3x + e-2x.

f'(x) = 3e3x - 2e-2x
I set f'(x) = 0 to find the critical numbers:
3e3x = 2e-2x
3 ln e3x = 2 ln e-2x
9x = -4x
x = 0, which is obviously wrong, (3e^0 - 2e^0 = 1). I found out that I had to combine the two terms using a common denomintor, and I got the right answer.

But why did my original method fail?
 
Physics news on Phys.org
Quite simply because;
3e3x = 2e-2x
3 ln e3x = 2 ln e-2x
[tex]3\ln|e^{3x}| \neq \ln |3e^{3x}|[/tex]
 
ahh... i feel so stupid.. i hate when i make simple mistakes and mix formulas up. :rolleyes:
 
endeavor said:
ahh... i feel so stupid.. i hate when i make simple mistakes and mix formulas up. :rolleyes:
I did a very similar thing in an exam yesterday, I was sat there for fifteen mintues trying to figure out where I had gone wrong, then I realized, I had differentiated instead of integrated :blushing:, it was a good job I realized before the end of the test! :rolleyes:
 

Similar threads

Replies
10
Views
2K
Replies
10
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
Replies
11
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K