Consider the function below
http://prntscr.com/90hzrg

- amy0799

Consider the function below
http://prntscr.com/90hzrg

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- 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

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

-9/4 and 0

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## More answers

- jango_IN_DTOWN

correct

- jango_IN_DTOWN

second part : the function is increasing when the first derivative is greater than 0

- jango_IN_DTOWN

find f'(x)

- amy0799

\[-\frac{ 2x+9 }{ 4x^3 }\]

- jango_IN_DTOWN

so this is >0 if 2x+9>0 i.e. if x> ???

- amy0799

-9/2, my bad

- jango_IN_DTOWN

so the interval is (-9/2,INFINITY)

- jango_IN_DTOWN

we are done with the second part

- jango_IN_DTOWN

third part see for decreasing f'(x)<0
we have already calculated f'(x)
so x<-9/2

- jango_IN_DTOWN

oh wait a bit

- jango_IN_DTOWN

-(2x+9)<0

- anonymous

A plot is attached.

##### 1 Attachment

- 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

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

how did you figure out the intervals?

- jango_IN_DTOWN

see second part we did a mistake, the function is increasing in (-9/2,0)

- jango_IN_DTOWN

ok see
x can be >0 or <0 right? @amy0799

- amy0799

right

- jango_IN_DTOWN

since x cannot be zero, or else the function is undefined

- jango_IN_DTOWN

now check the increasing part

- jango_IN_DTOWN

f'(X)>0
so
\[-(2x+9)/x^3>0\]

- 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

but it is not possible

- amy0799

i thought it's 4x^3

- jango_IN_DTOWN

4doesnot hamper the problem

- amy0799

oh ok

- 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

then x^3<0 and so
2x+9>0 , since the expression is >0

- jango_IN_DTOWN

from here we get x>-9/2 so the function is increasing in the interval (-9/2,0)

- jango_IN_DTOWN

@amy0799 we are done with the second and third part as well

- amy0799

It's decreasing at (0, infinity) and (-9/2,0)?

- jango_IN_DTOWN

correct

- amy0799

it's decreasing and increasing at the same interval?

- amy0799

at (-9/2,0)

- jango_IN_DTOWN

wait a bit

- jango_IN_DTOWN

no the decreasing intervals are (0,infinity) and (-infinity,-9/2)

- 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

@amy0799

- amy0799

oh ok, that makes sense

- jango_IN_DTOWN

now lets move to the final part..

- anonymous

Another plot to the left of the y axis

##### 1 Attachment

- 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

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

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

@amy0799 this completes the problem

- amy0799

is the relative extremum (-9/2, -0.056)?

- jango_IN_DTOWN

-9/2 and 0

- amy0799

it needs to be a point, so (-9/2,0)?

- jango_IN_DTOWN

why are you including braces? it will be -9/2 , 0

- amy0799

if you look at my attachment, it includes braces

- jango_IN_DTOWN

but it should be points.. ok.. you answer it (-9/2,0)

- 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

yeah those are -9/4 and 0

- jango_IN_DTOWN

wait the critical points are given by f'(X)=0 or undefined.
so the answers are -9/2 and 0

- 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

how do i find the y value of the relative extremum?
(-9/2, )

- 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

-0.056

- jango_IN_DTOWN

then (-9/2,-0.056) is the answer

- anonymous

There is an inflection point at
\[\left\{-\frac{27}{4},-\frac{4}{81}\right\} \]

- amy0799

i havent learned inflection

- anonymous

https://en.wikipedia.org/wiki/Inflection_point

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