need help finding intervals of increse....max and min...intervals of concavity up and down...and point of inflection
f '(x)=3e^x/(3+e^x)^2

- anonymous

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- amistre64

so this is our y' right?

- amistre64

e^x never equals zero, so there are no critical points

- anonymous

ok but i get an interval of increase right from some number to infinity

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## More answers

- amistre64

I would say yes, but to check for concavity we tend to use the 2nd derivative right?

- anonymous

well i know for sure that my function is increasing so there has to be interval of increasing but i dont know the starting point

- amistre64

e^x is never a negative number, so all of its positive right?

- anonymous

figured it out it from -infinity to infinity for increase variables ok now for concavity...

- amistre64

concave up is my guess.... but use 2nd derivative to be certain

- anonymous

i meant increase interval is -infinity to infinity

- amistre64

(3e^x) (3+e^x)^-2
-6 e^2x (3e^x)
--------- + ----------- right?
(3+e^x)^3 (3+e^x)^2

- amistre64

3 e^2x
---------
(3+e^x)^3

- amistre64

its always concave up....

- anonymous

my sources say otherwise and i am looking for the intervals from which it is concave up and down

- amistre64

i mighta goofed on the 2nd integral.... better double check

- anonymous

\[f"(x)=-(3e^{2x}-9e^x)/e ^{3x}+9e ^{2x}+27e^x+26\]

- amistre64

3e^x( e^3x -2e^2x +9e^x )
------------------------
(3+e^x)^4

- anonymous

no it not to the fourth power...

- amistre64

havent expanded the bottom to see if anything cancels, but thats just a left over from the quotient rule

- amistre64

it either that, or my heads got all jumbled..... which is very likely :)

- anonymous

oh no i think things cancel i am just saying the function i have is the correct function already checked it ....lol

- amistre64

my top gets to be:
-3e^x (e^x -1) i cant get it to match yours

- amistre64

-3e^x (e^x -3)

- amistre64

now it does lol

- anonymous

yea there you go so x=ln 3

- amistre64

yeah, when e^x = 3
x = ln(3) thats right... we get an inflection right?

- amistre64

everything to the left of that would be (+) and everything to the right of it would be(-)

- anonymous

yea...problem one complete now for two....consider the function f(x)=2x+4/3x+2 for this function there are two important intervals (-infinity, A) and (A, infinity) where the function is not defined at A. Find A

- amistre64

leftside is concave up, and rightside of it os concave down

- amistre64

A = -2/3 there is a vertical asymptote there and possess no real value.

- amistre64

nothing calcels top to bottom, so it aint a hole; its just the place where the denominator goes to zero... -2/3

- anonymous

ok yea that was correct....an i see you clarification

- anonymous

problem 3 sin^2(x/5) defined at interval [-14.907963,3.226991]

- anonymous

find where f(x) is concave down, global mini of the function, a local max which is not a global max, the function is increasing on a region....note: some answers must be given in interval notation

- amistre64

the normal period for pi is about 6 and this stretches it past 6 and goes to 10pi for the period. 10pi is about 31.4

- amistre64

the amplitude is exaggerated, but thats just the y value....

- amistre64

sin normally has a max at pi/2 and a min at 3pi/2
so, this one will be at 5pi/2 and 15pi/2 if I am picturing it correctly

- amistre64

but lets work it :)

- amistre64

Dx (3 sin^2(x/5))
y' = (3/5) sin(x/5)cos(x/5)

- amistre64

if sin or cos are zero, we get zeros...

- anonymous

the fx is sin^2(x/5) sorry its problem 3....

- amistre64

sin = 0 at 0 and 180, cos = 0 at 90 and 270

- amistre64

replace (3/5) with (1/5) same setup tho

- amistre64

but because the normal period is elongated to 10pi, we gotta find those numbers :)

- amistre64

they are..... 0, 5pi, 5pi/2, 15pi/2

- amistre64

we can go backwards to...
-5pi, -5pi/2, -15pi/2, -10pi

- amistre64

which of these numbers is in our interval?

- amistre64

5pi is bigger than 15

- amistre64

[-14.907963 , 3.226991]
-15/2 = -7.5, so -5pi/2 is in our interval
0 is in our interval
we got these 2 numbers (-5pi/2) and (0)

- anonymous

are you using \[\sin ^2(x/5)\]

- amistre64

yes, I am using the derivative of that and checking for zeros :)

- amistre64

the 3 was superfluous and could be tossed on a whim...

- anonymous

ok

- anonymous

ok the global mini is at 0 but now we need a local max

- amistre64

-5pi/2 would be the local max...

- anonymous

nah it not...i checked

- amistre64

this is what we look like I think

##### 1 Attachment

- anonymous

our local max was 3.22 from the interval

- anonymous

i need the interval where it is concave down and region where it is increasing

- amistre64

computer froze

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