1. Michele_Laino

here we have to compute the first derivative of your function

2. Michele_Laino

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

3. Michele_Laino

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

4. 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?

5. anonymous

e^x>1/2

6. Michele_Laino

please now solve it for x

7. anonymous

would it be ln(e)>1/2

8. Michele_Laino

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

9. Michele_Laino

am I right?

10. Michele_Laino

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

11. anonymous

oh okay i see it now

12. Michele_Laino

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

13. Michele_Laino

the domain of your function is all the real line

14. 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?

15. Michele_Laino

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

16. Michele_Laino

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

17. Michele_Laino

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

18. anonymous

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

19. 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: $\Large f\left( { - \ln 2} \right) = \frac{1}{4} - \frac{1}{2} = - \frac{1}{4}$

20. anonymous

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

21. Michele_Laino

it is a local minimum

22. anonymous

wait thats the min right?

23. Michele_Laino

right!

24. anonymous

and theres no max for this function right?

25. Michele_Laino

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

26. 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)?

27. Michele_Laino

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

28. Michele_Laino

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

29. Michele_Laino

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

30. Michele_Laino

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

31. 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?

32. anonymous

would it be x>ln1/4

33. Michele_Laino

that's right!

34. Michele_Laino

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

35. Michele_Laino

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

36. 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?

37. anonymous

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

38. Michele_Laino

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

39. Michele_Laino

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

40. anonymous

oh okay i understand now

41. anonymous

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

42. Michele_Laino

:)