Question

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

Answer 1

critical numbers imply the value of x for which the given function takes value 0 or is undefined, i.e. the denominator is zero

Answer 2

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??

Answer 3

-9/4 and 0

Answer 4

correct

Answer 5

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

Answer 6

find f'(x)

Answer 7

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

Answer 8

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

Answer 9

-9/2, my bad

Answer 10

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

Answer 11

we are done with the second part

Answer 12

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

Answer 13

oh wait a bit

Answer 14

-(2x+9)<0

Answer 15

A plot is attached.

Answer 16

-(2x+9)/x^3<0 so here we have to check the va;ues of the denominator as well

Answer 17

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)

Answer 18

how did you figure out the intervals?

Answer 19

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

Answer 20

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

Answer 21

right

Answer 22

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

Answer 23

now check the increasing part

Answer 24

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

Answer 25

first see if x>0, then 2x+9>0 and x^3>0 so the expression
\[(2x+9)/x^3>0\]

Answer 26

but it is not possible

Answer 27

i thought it's 4x^3

Answer 28

4doesnot hamper the problem

Answer 29

oh ok

Answer 30

since x>0 is not possible as we figured out, so we are left with the possibility that x<0

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

correct

Answer 36

it's decreasing and increasing at the same interval?

Answer 37

at (-9/2,0)

Answer 38

wait a bit

Answer 39

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

Answer 40

just consider 2 cases x>0 and x<0.. in both the cases you can figure out the intervals

Answer 41

@amy0799

Answer 42

oh ok, that makes sense

Answer 43

now lets move to the final part..

Answer 44

Another plot to the left of the y axis

Answer 45

see the increasing interval is (-9/2,0)
and the decreasing intervals are (-infinity,-9/2) and (0, infinity)

Answer 46

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

Answer 47

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

Answer 48

@amy0799 this completes the problem

Answer 49

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

Answer 50

-9/2 and 0

Answer 51

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

Answer 52

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

Answer 53

if you look at my attachment, it includes braces

Answer 54

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

Answer 55

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

Answer 56

yeah those are -9/4 and 0

Answer 57

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

Answer 58

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.

Answer 59

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

Answer 60

for this value of x, fund the value of y by plugging in the value of x in the expression of y

Answer 61

-0.056

Answer 62

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

Answer 63

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

Answer 64

i havent learned inflection

Answer 65

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