Question

Find the interval of increase and the interval of decrease of 200+8x^3+x^4

Answer 1

did you do first derivative?

Answer 2

\[(200+8x^3+x^4)'=24x^2+4x^3=4x^2(6+x)\]

Answer 3

set that equal zero for critical points

Answer 4

@Gaby7890 ?

Answer 5

I forgot to put by using the second derivative test

Answer 6

second derivative shows concavity of the curve

Answer 7

here you just need interval of increase and decrease

Answer 8

Yes, I need the intervals where it increase and the intervals where it decreases by using the second derivative test.

Answer 9

\(4x^2(6+x)=0 \Longrightarrow x=0 ~~or~~ x=-6\)
critical points at x=0 and x=-6

Answer 10

now let's see \[\text{this expression } 4x^2(x+6) ~~\text{the sign here depends on} ~~(x+6)~~ \text{because }~~ 4x^2>0 ~~\text{for any x} \]

Answer 11

Well my critical points are at X=0 and X=-4 by finding the Second derivative of 200+8x^3+x^4

Answer 12

hmm how come you get x=-4? can you show the work

Answer 13

as i said there no need for second derivative test
second derivative test is used to show how the graph is curving concave up or down and show inflection point

Answer 14

you don't need second derivative

Answer 15

f(X)=200+8x^3+x^4
f'(X)=24x^2+4x^3
f''(X)=48x^2+12x^2
=12x(X+4)

Answer 16

so now we know that f'(x)<0 fo x<-6 and f(x)>0 for x>-6

Answer 17

correction f'(x)>0 for x>-6

Answer 18

X=0 and X=-4 are my inflection points

Answer 19

x=-4 is the where the inflection point happens

Answer 20

yes that's true

Answer 21

but your question is asking interval of increase and decrease not inflection point

Answer 22

I'm trying to find the the intervals where it increases

Answer 23

you are doing an unnecessary work

Answer 24

i just did it we have f'<0 for x<-6
f'>0 for x>-6
meaning f is increasing for x>-6 and decreasing for x<-6

Answer 25

here is the graph that shows same result i got
http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427e8jc7pjgfe5

Answer 26

Okay thank you for the help. By any chance do you know how to do 4csc(3x/2) with the intervals (0,2pi)?

Answer 27

as you can see at x=0 and at x=-4 inflection points as you said exactly

Answer 28

same principle

Answer 29

Okay but this is another problem I need help with.

Answer 30

\[(4\csc(3x/2))'=-6\csc(3x/2)\tan(3x/2)\]

Answer 31

now you just look at the sign of tan(3x/2) for x in [0,2pi)
tan is positive for [0,pi/2) and [pi, 3pi/2)
for csc which is just 1/sin
positive in first quadrant and second quadrant and negative in III, IV

Answer 32

you just find then the sign of the product

Answer 33

Would trying to find the second derivative of 4csc(3x/2) change anything. Because I have to find the (leftmost, interval, smaller X value) and also I have to find the (leftmost interval, larger X value)?

Answer 34

By using the second derivative test.

Answer 35

hmm i don't know what you mean by leftmost? any example

Answer 36

oh correction derivative should have cotangent not tangent sorry blind it there
i was thinking somehow of sec instead of csc

Answer 37

should be -6cot(3x/2)csc(3x/2) correction
but the my talk about signs still holds since cot is just 1/tan

Answer 38

Consider the function below on the interval (0, 2π). (Give your answers correct to 2 decimal places.)
f(x) = 4 csc((3x)/2)
(a) Find the points of inflection of the function. (If none exist, enter NONE.)



(b) Find the open intervals where the graph is concave upward.
(
Correct: Your answer is correct.
,
Incorrect: Your answer is incorrect.
) (leftmost interval, smaller x value)
( ,
Correct: Your answer is correct.
) (rightmost interval, larger x value)

(c) Find the open interval where the graph is concave downward.
(
Incorrect: Your answer is incorrect.
,
Incorrect: Your answer is incorrect.
)

Answer 39

oh yeah you definitely need second derivative for this one
it is asking for inflection points
those leftmost and rightmost
they are just asking where it is concaved up
that happens in two interval here
that's why they give you two
and then last question where it is concaved down

Answer 40

here we don't need first derivative to answer this
we just do it and skip it and derive second derivative to answer this problem

Answer 41

it needs some more work to clean!
i gotta go anyway to sleep

Answer 42

good luck weetie!

Answer 43

Okay thank you!

Answer 44

no problem