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

- anonymous

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

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- Michele_Laino

here we have to compute the first derivative of your function

- Michele_Laino

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

- Michele_Laino

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- Michele_Laino

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?

- anonymous

e^x>1/2

- Michele_Laino

please now solve it for x

- anonymous

would it be ln(e)>1/2

- Michele_Laino

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

- Michele_Laino

am I right?

- Michele_Laino

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

- anonymous

oh okay i see it now

- Michele_Laino

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

- Michele_Laino

the domain of your function is all the real line

- anonymous

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

- Michele_Laino

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

- Michele_Laino

now we have this:
|dw:1435213594909:dw|

- Michele_Laino

as we can see, point x=-ln(2) is a local minimum for f(x)

- anonymous

and I plug that into the first derivative so i can get the interval?

- Michele_Laino

you can get the intervals in which your function is an increasing or a decreasing function, only solving the inequality:
f '(x) >0 for example
Now we have found
x=-ln(2) or, equivalently e^x= 1/2.
At that point the value of your function is:
\[\Large f\left( { - \ln 2} \right) = \frac{1}{4} - \frac{1}{2} = - \frac{1}{4}\]

- anonymous

so my max would be at (-ln(2),-1/4)?

- Michele_Laino

it is a local minimum

- anonymous

wait thats the min right?

- Michele_Laino

right!

- anonymous

and theres no max for this function right?

- Michele_Laino

that's right! there are no points of local maximum

- anonymous

for my concave up and concave down, would it be for concave down (-ln(2),-1/4)
and for concave up (-infinity,-ln(2))U(-ln(2),+infinity)?

- Michele_Laino

in order to establish the concavity type, we have to study the sign of the second derivative

- Michele_Laino

after a simple computation, I got this:
\[\large f''\left( x \right) = {e^x}\left( {4{e^x} - 1} \right)\]

- Michele_Laino

now, our function is concave up when f ''(x)>0, and it is concave down when f ''(x)<0
|dw:1435214691996:dw|

- Michele_Laino

so we have to solve this inequality:
\[\Large f''\left( x \right) > 0 \Rightarrow {e^x}\left( {4{e^x} - 1} \right) > 0\]

- Michele_Laino

next, since e^x is always positive, then that inequality is equivalent to this one:
\[\Large 4{e^x} - 1 > 0\]
please solve it for x, what do you get?

- anonymous

would it be x>ln1/4

- Michele_Laino

that's right!

- Michele_Laino

we can write it as below:
\[\Large x > - \ln 4\]

- Michele_Laino

so your function, is concave down in this interval:
(-infinity, -ln4),
and concave up in this interval:
[-ln4, +infinity)

- anonymous

to find the inflection point I would plug in -ln4 into the second derivative and then that result back into the original equation right?

- anonymous

oh wait nvm i would solve for x in the second derivative?

- Michele_Laino

no, you have to plug in -ln4 into the function, like this:
\[\large f\left( { - \ln 4} \right) = {\left( {\frac{1}{4}} \right)^2} - \frac{1}{4} = \frac{1}{{16}} - \frac{1}{4} = - \frac{3}{{16}}\]

- Michele_Laino

so your inflection point is:
\[\Large \left( { - \ln 4, - \frac{3}{{16}}} \right)\]

- anonymous

oh okay i understand now

- anonymous

thank you so much!! I really appreciated it!!

- Michele_Laino

:)

Looking for something else?

Not the answer you are looking for? Search for more explanations.