anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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amistre64
  • amistre64
so this is our y' right?
amistre64
  • amistre64
e^x never equals zero, so there are no critical points
anonymous
  • anonymous
ok but i get an interval of increase right from some number to infinity

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amistre64
  • amistre64
I would say yes, but to check for concavity we tend to use the 2nd derivative right?
anonymous
  • 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
  • amistre64
e^x is never a negative number, so all of its positive right?
anonymous
  • anonymous
figured it out it from -infinity to infinity for increase variables ok now for concavity...
amistre64
  • amistre64
concave up is my guess.... but use 2nd derivative to be certain
anonymous
  • anonymous
i meant increase interval is -infinity to infinity
amistre64
  • amistre64
(3e^x) (3+e^x)^-2 -6 e^2x (3e^x) --------- + ----------- right? (3+e^x)^3 (3+e^x)^2
amistre64
  • amistre64
3 e^2x --------- (3+e^x)^3
amistre64
  • amistre64
its always concave up....
anonymous
  • anonymous
my sources say otherwise and i am looking for the intervals from which it is concave up and down
amistre64
  • amistre64
i mighta goofed on the 2nd integral.... better double check
anonymous
  • anonymous
\[f"(x)=-(3e^{2x}-9e^x)/e ^{3x}+9e ^{2x}+27e^x+26\]
amistre64
  • amistre64
3e^x( e^3x -2e^2x +9e^x ) ------------------------ (3+e^x)^4
anonymous
  • anonymous
no it not to the fourth power...
amistre64
  • amistre64
havent expanded the bottom to see if anything cancels, but thats just a left over from the quotient rule
amistre64
  • amistre64
it either that, or my heads got all jumbled..... which is very likely :)
anonymous
  • anonymous
oh no i think things cancel i am just saying the function i have is the correct function already checked it ....lol
amistre64
  • amistre64
my top gets to be: -3e^x (e^x -1) i cant get it to match yours
amistre64
  • amistre64
-3e^x (e^x -3)
amistre64
  • amistre64
now it does lol
anonymous
  • anonymous
yea there you go so x=ln 3
amistre64
  • amistre64
yeah, when e^x = 3 x = ln(3) thats right... we get an inflection right?
amistre64
  • amistre64
everything to the left of that would be (+) and everything to the right of it would be(-)
anonymous
  • 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
  • amistre64
leftside is concave up, and rightside of it os concave down
amistre64
  • amistre64
A = -2/3 there is a vertical asymptote there and possess no real value.
amistre64
  • amistre64
nothing calcels top to bottom, so it aint a hole; its just the place where the denominator goes to zero... -2/3
anonymous
  • anonymous
ok yea that was correct....an i see you clarification
anonymous
  • anonymous
problem 3 sin^2(x/5) defined at interval [-14.907963,3.226991]
anonymous
  • 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
  • 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
  • amistre64
the amplitude is exaggerated, but thats just the y value....
amistre64
  • 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
  • amistre64
but lets work it :)
amistre64
  • amistre64
Dx (3 sin^2(x/5)) y' = (3/5) sin(x/5)cos(x/5)
amistre64
  • amistre64
if sin or cos are zero, we get zeros...
anonymous
  • anonymous
the fx is sin^2(x/5) sorry its problem 3....
amistre64
  • amistre64
sin = 0 at 0 and 180, cos = 0 at 90 and 270
amistre64
  • amistre64
replace (3/5) with (1/5) same setup tho
amistre64
  • amistre64
but because the normal period is elongated to 10pi, we gotta find those numbers :)
amistre64
  • amistre64
they are..... 0, 5pi, 5pi/2, 15pi/2
amistre64
  • amistre64
we can go backwards to... -5pi, -5pi/2, -15pi/2, -10pi
amistre64
  • amistre64
which of these numbers is in our interval?
amistre64
  • amistre64
5pi is bigger than 15
amistre64
  • 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
  • anonymous
are you using \[\sin ^2(x/5)\]
amistre64
  • amistre64
yes, I am using the derivative of that and checking for zeros :)
amistre64
  • amistre64
the 3 was superfluous and could be tossed on a whim...
anonymous
  • anonymous
ok
anonymous
  • anonymous
ok the global mini is at 0 but now we need a local max
amistre64
  • amistre64
-5pi/2 would be the local max...
anonymous
  • anonymous
nah it not...i checked
amistre64
  • amistre64
this is what we look like I think
1 Attachment
anonymous
  • anonymous
our local max was 3.22 from the interval
anonymous
  • anonymous
i need the interval where it is concave down and region where it is increasing
amistre64
  • amistre64
computer froze

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