anonymous
  • anonymous
find the intervals of increase and decrease for e^2x-e^x, please help!
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Michele_Laino
  • Michele_Laino
here we have to compute the first derivative of your function
Michele_Laino
  • Michele_Laino
I got this: \[\Large f'\left( x \right) = {e^x}\left( {2{e^x} - 1} \right)\]
Michele_Laino
  • Michele_Laino
now the intervals of the real line in which f(x) is increasing, are given by the subsequent condition: \[\Large f'\left( x \right) > 0 \Rightarrow {e^x}\left( {2{e^x} - 1} \right) > 0\]

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Michele_Laino
  • 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
  • anonymous
e^x>1/2
Michele_Laino
  • Michele_Laino
please now solve it for x
anonymous
  • anonymous
would it be ln(e)>1/2
Michele_Laino
  • 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
  • Michele_Laino
am I right?
Michele_Laino
  • 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
  • anonymous
oh okay i see it now
Michele_Laino
  • Michele_Laino
so your function is increasing in: [-ln 2, +infinity) and decreasing, in: (-infinity, -ln2)
Michele_Laino
  • Michele_Laino
the domain of your function is all the real line
anonymous
  • 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
  • Michele_Laino
no, the local max and min comes from the first derivative
Michele_Laino
  • Michele_Laino
now we have this: |dw:1435213594909:dw|
Michele_Laino
  • Michele_Laino
as we can see, point x=-ln(2) is a local minimum for f(x)
anonymous
  • anonymous
and I plug that into the first derivative so i can get the interval?
Michele_Laino
  • 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
  • anonymous
so my max would be at (-ln(2),-1/4)?
Michele_Laino
  • Michele_Laino
it is a local minimum
anonymous
  • anonymous
wait thats the min right?
Michele_Laino
  • Michele_Laino
right!
anonymous
  • anonymous
and theres no max for this function right?
Michele_Laino
  • Michele_Laino
that's right! there are no points of local maximum
anonymous
  • 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
  • Michele_Laino
in order to establish the concavity type, we have to study the sign of the second derivative
Michele_Laino
  • Michele_Laino
after a simple computation, I got this: \[\large f''\left( x \right) = {e^x}\left( {4{e^x} - 1} \right)\]
Michele_Laino
  • 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
  • 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
  • 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
  • anonymous
would it be x>ln1/4
Michele_Laino
  • Michele_Laino
that's right!
Michele_Laino
  • Michele_Laino
we can write it as below: \[\Large x > - \ln 4\]
Michele_Laino
  • Michele_Laino
so your function, is concave down in this interval: (-infinity, -ln4), and concave up in this interval: [-ln4, +infinity)
anonymous
  • 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
  • anonymous
oh wait nvm i would solve for x in the second derivative?
Michele_Laino
  • 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
  • Michele_Laino
so your inflection point is: \[\Large \left( { - \ln 4, - \frac{3}{{16}}} \right)\]
anonymous
  • anonymous
oh okay i understand now
anonymous
  • anonymous
thank you so much!! I really appreciated it!!
Michele_Laino
  • Michele_Laino
:)

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