Question

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

Answer 1

so this is our y' right?

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

i meant increase interval is -infinity to infinity

Answer 10

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

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

Answer 11

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

Answer 12

its always concave up....

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

no it not to the fourth power...

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

my top gets to be:

-3e^x (e^x -1) i cant get it to match yours

Answer 22

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

Answer 23

now it does lol

Answer 24

yea there you go so x=ln 3

Answer 25

yeah, when e^x = 3

x = ln(3) thats right... we get an inflection right?

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

but lets work it :)

Answer 38

Dx (3 sin^2(x/5))

y' = (3/5) sin(x/5)cos(x/5)

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

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

Answer 43

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

Answer 44

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

Answer 45

we can go backwards to...

-5pi, -5pi/2, -15pi/2, -10pi

Answer 46

which of these numbers is in our interval?

Answer 47

5pi is bigger than 15

Answer 48

[-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)

Answer 49

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

Answer 50

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

Answer 51

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

Answer 52

ok

Answer 53

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

Answer 54

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

Answer 55

nah it not...i checked

Answer 56

this is what we look like I think

Answer 57

our local max was 3.22 from the interval

Answer 58

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

Answer 59

computer froze