amy0799
  • amy0799
Consider the function below http://prntscr.com/90hzrg
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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jango_IN_DTOWN
  • jango_IN_DTOWN
critical numbers imply the value of x for which the given function takes value 0 or is undefined, i.e. the denominator is zero
jango_IN_DTOWN
  • jango_IN_DTOWN
numerator =0 implies 4x+9=0 and from here calculate the x value denominator =0 implies x^2=0 implies x=0 so you get two critical numbers which are??
amy0799
  • amy0799
-9/4 and 0

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jango_IN_DTOWN
  • jango_IN_DTOWN
correct
jango_IN_DTOWN
  • jango_IN_DTOWN
second part : the function is increasing when the first derivative is greater than 0
jango_IN_DTOWN
  • jango_IN_DTOWN
find f'(x)
amy0799
  • amy0799
\[-\frac{ 2x+9 }{ 4x^3 }\]
jango_IN_DTOWN
  • jango_IN_DTOWN
so this is >0 if 2x+9>0 i.e. if x> ???
amy0799
  • amy0799
-9/2, my bad
jango_IN_DTOWN
  • jango_IN_DTOWN
so the interval is (-9/2,INFINITY)
jango_IN_DTOWN
  • jango_IN_DTOWN
we are done with the second part
jango_IN_DTOWN
  • jango_IN_DTOWN
third part see for decreasing f'(x)<0 we have already calculated f'(x) so x<-9/2
jango_IN_DTOWN
  • jango_IN_DTOWN
oh wait a bit
jango_IN_DTOWN
  • jango_IN_DTOWN
-(2x+9)<0
anonymous
  • anonymous
A plot is attached.
1 Attachment
jango_IN_DTOWN
  • jango_IN_DTOWN
-(2x+9)/x^3<0 so here we have to check the va;ues of the denominator as well
jango_IN_DTOWN
  • jango_IN_DTOWN
case 1) when x>0 then -(2x+9)<0 implies 2x+9>0 implies x>-9/2 case 2) when x<0 then 2x+9>0 implies x>-9/2 so we get two intervals (0,infinity) and (-9/2,0)
amy0799
  • amy0799
how did you figure out the intervals?
jango_IN_DTOWN
  • jango_IN_DTOWN
see second part we did a mistake, the function is increasing in (-9/2,0)
jango_IN_DTOWN
  • jango_IN_DTOWN
ok see x can be >0 or <0 right? @amy0799
amy0799
  • amy0799
right
jango_IN_DTOWN
  • jango_IN_DTOWN
since x cannot be zero, or else the function is undefined
jango_IN_DTOWN
  • jango_IN_DTOWN
now check the increasing part
jango_IN_DTOWN
  • jango_IN_DTOWN
f'(X)>0 so \[-(2x+9)/x^3>0\]
jango_IN_DTOWN
  • jango_IN_DTOWN
first see if x>0, then 2x+9>0 and x^3>0 so the expression \[(2x+9)/x^3>0\]
jango_IN_DTOWN
  • jango_IN_DTOWN
but it is not possible
amy0799
  • amy0799
i thought it's 4x^3
jango_IN_DTOWN
  • jango_IN_DTOWN
4doesnot hamper the problem
amy0799
  • amy0799
oh ok
jango_IN_DTOWN
  • jango_IN_DTOWN
since x>0 is not possible as we figured out, so we are left with the possibility that x<0
jango_IN_DTOWN
  • jango_IN_DTOWN
then x^3<0 and so 2x+9>0 , since the expression is >0
jango_IN_DTOWN
  • jango_IN_DTOWN
from here we get x>-9/2 so the function is increasing in the interval (-9/2,0)
jango_IN_DTOWN
  • jango_IN_DTOWN
@amy0799 we are done with the second and third part as well
amy0799
  • amy0799
It's decreasing at (0, infinity) and (-9/2,0)?
jango_IN_DTOWN
  • jango_IN_DTOWN
correct
amy0799
  • amy0799
it's decreasing and increasing at the same interval?
amy0799
  • amy0799
at (-9/2,0)
jango_IN_DTOWN
  • jango_IN_DTOWN
wait a bit
jango_IN_DTOWN
  • jango_IN_DTOWN
no the decreasing intervals are (0,infinity) and (-infinity,-9/2)
jango_IN_DTOWN
  • jango_IN_DTOWN
just consider 2 cases x>0 and x<0.. in both the cases you can figure out the intervals
jango_IN_DTOWN
  • jango_IN_DTOWN
@amy0799
amy0799
  • amy0799
oh ok, that makes sense
jango_IN_DTOWN
  • jango_IN_DTOWN
now lets move to the final part..
anonymous
  • anonymous
Another plot to the left of the y axis
1 Attachment
jango_IN_DTOWN
  • jango_IN_DTOWN
see the increasing interval is (-9/2,0) and the decreasing intervals are (-infinity,-9/2) and (0, infinity)
jango_IN_DTOWN
  • jango_IN_DTOWN
so to the left of x=-9/2, the function is decreasing at to the right of x=-9/2, it is increasing, so x=-9/2 is a point of relative minimum
jango_IN_DTOWN
  • jango_IN_DTOWN
and at the left of x=0. teh function is increasing and to the right of x=0 it is decreaing. So we have relative maximum at x=0
jango_IN_DTOWN
  • jango_IN_DTOWN
@amy0799 this completes the problem
amy0799
  • amy0799
is the relative extremum (-9/2, -0.056)?
jango_IN_DTOWN
  • jango_IN_DTOWN
-9/2 and 0
amy0799
  • amy0799
it needs to be a point, so (-9/2,0)?
jango_IN_DTOWN
  • jango_IN_DTOWN
why are you including braces? it will be -9/2 , 0
amy0799
  • amy0799
if you look at my attachment, it includes braces
jango_IN_DTOWN
  • jango_IN_DTOWN
but it should be points.. ok.. you answer it (-9/2,0)
amy0799
  • amy0799
i got the relative extremum wrong, the -9/2 is correct and for the critical points, i need the number, not how many critical points there are
jango_IN_DTOWN
  • jango_IN_DTOWN
yeah those are -9/4 and 0
jango_IN_DTOWN
  • jango_IN_DTOWN
wait the critical points are given by f'(X)=0 or undefined. so the answers are -9/2 and 0
anonymous
  • anonymous
The curve begins to move upwards at x = -(27/4) or -6.75 To confirm, set the 2nd derivative to zero and then solve for x.
amy0799
  • amy0799
how do i find the y value of the relative extremum? (-9/2, )
jango_IN_DTOWN
  • jango_IN_DTOWN
for this value of x, fund the value of y by plugging in the value of x in the expression of y
amy0799
  • amy0799
-0.056
jango_IN_DTOWN
  • jango_IN_DTOWN
then (-9/2,-0.056) is the answer
anonymous
  • anonymous
There is an inflection point at \[\left\{-\frac{27}{4},-\frac{4}{81}\right\} \]
amy0799
  • amy0799
i havent learned inflection
anonymous
  • anonymous
https://en.wikipedia.org/wiki/Inflection_point

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