Question

Find the interval on which the function is increasing or decreasing

Answer 1

\[f(x)=e^{2x}+e^{-x}\]

Answer 2

\[f'x=2e^{2x}-e^{-x}\]

Answer 3

Hello c:

Answer 4

yes go on..f'x > 0 for increasing and < 0 for decreasing

Answer 5

\[e^{-x}(2e^{2x}-1)\]

Answer 6

so when i set it equal to zero,

Answer 7

\[2e^{2x}-1=0\]

Answer 8

\[e^{2x}=1/2\]

Answer 9

i hate e lol

Answer 10

or would it be easier to do this

Answer 11

\[f'x=2e^{2x}-e^{-x}=0\]

Answer 12

\[2e^{2x}=e^{-x}\]

Answer 13

Yes it is e^2x > 1/2 for the function to be increasing
Apply logarithm on both sides,
2x > log(1/2)
2x > -log2
x > (-log2)/2

Answer 14

Both the ways, you get the same. Actually, your first method is relatively easier

Answer 15

Now put the symbol < in the place of > to get the interval for decreasing function

Answer 16

ok the book is getting -1/3ln2, did i differentiate wrong?

Answer 17

also how did you get that? e^2x>1/2

Answer 18

Yes your book is right..it is e^3x > 1/2 and then proceed as it is by replacing 2 by 3 in the above solution.

Answer 19

\[e^{3x}=\frac{ 1 }{ 2 }\]

Answer 20

where i get lost

Answer 21

After you differentiated the function and took out e^-x as a common term, you got this:
f'x = e^-x(e^3x - 1)
You did it yourself above and its correct.
Now, f(x) is increasing when f'(x) > o (not equal to zero)
It should be e^3x > 1/2 and not e^3x = 1/2

Answer 22

Similarly, f(x) is decreasing when f'(x) < 0 , i.e., when e^3x < 1/2

Answer 23

Did you get it?

Answer 24

yes, the part i dont get is the log part,

Answer 25

somehow e^3x>1/2 equals -1/3ln2

Answer 26

Did you undergo a course involving logarithms?

Answer 27

yes i will look it up

Answer 28

thanks

Answer 29

You are welcome