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anonymous
 5 years ago
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
 5 years ago
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

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0so this is our y' right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0e^x never equals zero, so there are no critical points

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok but i get an interval of increase right from some number to infinity

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0I would say yes, but to check for concavity we tend to use the 2nd derivative right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well 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
 5 years ago
Best ResponseYou've already chosen the best response.0e^x is never a negative number, so all of its positive right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0figured it out it from infinity to infinity for increase variables ok now for concavity...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0concave up is my guess.... but use 2nd derivative to be certain

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i meant increase interval is infinity to infinity

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0(3e^x) (3+e^x)^2 6 e^2x (3e^x)  +  right? (3+e^x)^3 (3+e^x)^2

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.03 e^2x  (3+e^x)^3

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0its always concave up....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0my sources say otherwise and i am looking for the intervals from which it is concave up and down

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i mighta goofed on the 2nd integral.... better double check

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[f"(x)=(3e^{2x}9e^x)/e ^{3x}+9e ^{2x}+27e^x+26\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.03e^x( e^3x 2e^2x +9e^x )  (3+e^x)^4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no it not to the fourth power...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0havent expanded the bottom to see if anything cancels, but thats just a left over from the quotient rule

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0it either that, or my heads got all jumbled..... which is very likely :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh no i think things cancel i am just saying the function i have is the correct function already checked it ....lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0my top gets to be: 3e^x (e^x 1) i cant get it to match yours

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea there you go so x=ln 3

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, when e^x = 3 x = ln(3) thats right... we get an inflection right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0everything to the left of that would be (+) and everything to the right of it would be()

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea...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
 5 years ago
Best ResponseYou've already chosen the best response.0leftside is concave up, and rightside of it os concave down

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0A = 2/3 there is a vertical asymptote there and possess no real value.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0nothing calcels top to bottom, so it aint a hole; its just the place where the denominator goes to zero... 2/3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok yea that was correct....an i see you clarification

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0problem 3 sin^2(x/5) defined at interval [14.907963,3.226991]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0find 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
 5 years ago
Best ResponseYou've already chosen the best response.0the 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
 5 years ago
Best ResponseYou've already chosen the best response.0the amplitude is exaggerated, but thats just the y value....

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0sin 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
 5 years ago
Best ResponseYou've already chosen the best response.0Dx (3 sin^2(x/5)) y' = (3/5) sin(x/5)cos(x/5)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if sin or cos are zero, we get zeros...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the fx is sin^2(x/5) sorry its problem 3....

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0sin = 0 at 0 and 180, cos = 0 at 90 and 270

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0replace (3/5) with (1/5) same setup tho

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0but because the normal period is elongated to 10pi, we gotta find those numbers :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0they are..... 0, 5pi, 5pi/2, 15pi/2

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we can go backwards to... 5pi, 5pi/2, 15pi/2, 10pi

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0which of these numbers is in our interval?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.05pi is bigger than 15

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0[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
 5 years ago
Best ResponseYou've already chosen the best response.0are you using \[\sin ^2(x/5)\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0yes, I am using the derivative of that and checking for zeros :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the 3 was superfluous and could be tossed on a whim...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok the global mini is at 0 but now we need a local max

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.05pi/2 would be the local max...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nah it not...i checked

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0this is what we look like I think

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0our local max was 3.22 from the interval

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i need the interval where it is concave down and region where it is increasing
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