Question

find the intervals of increase and decrease for e^2x-e^x, please help!

Answer 1

here we have to compute the first derivative of your function

Answer 2

I got this:
\[\Large f'\left( x \right) = {e^x}\left( {2{e^x} - 1} \right)\]

Answer 3

now the intervals of the real line in which f(x) is increasing, are given by the subsequent condition:
\[\Large f'\left( x \right) > 0 \Rightarrow {e^x}\left( {2{e^x} - 1} \right) > 0\]

Answer 4

which, since e^x is always positive, is equivalent to this one:
\[\Large 2{e^x} - 1 > 0\]
please solve that inequality, what do you get?

Answer 5

e^x>1/2

Answer 6

please now solve it for x

Answer 7

would it be ln(e)>1/2

Answer 8

no, since if we take the logarithm of both sides, we get:
\[\Large \begin{gathered}
x\left( {\ln e} \right) > \ln \left( {1/2} \right) \hfill \\
x > - \ln 2 \hfill \\
\end{gathered} \]

Answer 9

am I right?

Answer 10

please remember that:
\[\Large \begin{gathered}
\ln e = 1 \hfill \\
\ln \left( {1/2} \right) = \ln 1 - \ln 2 = 0 - \ln 2 = - \ln 2 \hfill \\
\end{gathered} \]

Answer 11

oh okay i see it now

Answer 12

so your function is increasing in:
[-ln 2, +infinity)
and decreasing, in:
(-infinity, -ln2)

Answer 13

the domain of your function is all the real line

Answer 14

okay! and to find the local max and min i would have to find the second derivative and plug in -ln2 for x right?

Answer 15

no, the local max and min comes from the first derivative

Answer 16

now we have this: