Question

x−6x^(1/3) domain pls

Answer 1

R

Answer 2

All real number R

Answer 3

oh wow.

what about the x and y-int

Answer 4

I dislike this kind of question, domain is part of function specification.

Answer 5

yeah me too.

Answer 6

can you help me with the x and y-int of this pls

Answer 7

sure

Answer 8

when you set x = 0 , you will get y = 0 as well

when you set y 0, you will get this equation which is x−6x^(1/3) = 0

To solve it:

x = 6 x^(1/3)

cube both sides

x^3 = 216 x

x^3 - 216x = 0

x (x^2 - 216) = 0

x = 0 OR x^2 -216=0

Answer 9

my software didnt accept the ans

Answer 10

hmmmmmmmm

Answer 11

Is this an equation or what?

Answer 12

@Jinnie

Answer 13

its a function

f(x)= x−6x^(1/3)

Answer 14

are you there

Answer 15

Can you try (0,0) and (14.69 , 0)

Answer 16

neither worked

Answer 17

So I don't know because it's complicated equation

Sorry to say that

Answer 18

ok what about this

Answer 19

?

Answer 20

(x)=3cosx−cos^(3)x on the interval 0<x<2π

Answer 21

whats the x-int for that

Answer 22

ok wait

Answer 23

Is it f(x) in the begining or just (x)

Answer 24

f(x)=

Answer 25

sry

Answer 26

ok np

Answer 27

it's gonna be pi/2

Answer 28

(pi/2 , 0)

Answer 29

right? @Jinnie

Answer 30

program says no

Answer 31

really?

Answer 32

oh sorry it's (pi/2 , 0) OR (3pi/2 , 0)

Answer 33

check these please @Jinnie

Answer 34

look at the doc attached

Answer 35

you wrote it in wrong order and what is the union sign between them?

Answer 36

Don't make union sigh between them
it's just a points

Answer 37

i tried (pi/2,0)

and


(3pi/2,0)

neither worked

Answer 38

I don't really know :(
I'm sorry about that

Answer 39

ok might you know the local maximum?

Answer 40

Actually I can't remember it
I'm sorry again

Answer 41

ok what about this

-9sin^2(x) cos(x)

can you tell which interval this is increasing and decreasing on

Answer 42

not increasing and decreasing but concave up and down

Answer 43

I think you have to take the derivative and then make it equal to 0 and find the values for x

Answer 44

Can you do that?

Answer 45

ive tried doing that but im not getting the right answer

Answer 46

-9sin^2(x) cos(x)<<<< this is already in derivative form

so -9sin^2(x) cos(x)= 0 and i get nonsense

Answer 47

I don't really remember this stuff
I'm sorry @Jinnie

Answer 48

ok np

Answer 49

wow this is a long thread. what is the actual question?
domain is all real numbers. after than i am lost as to what the question is

Answer 50

there are a couple but can we start with this pls

-9sin^2(x) cos(x)....where is it concave up and down....its already in second derivative form

Answer 51

\[f(x)=(x)=3\cos(x)−\cos^3(x)\]

Answer 52

what is the original function?

Answer 53

3cosx−cos^(3)x for 0<x<2π

Answer 54

ok got it

Answer 55

so first derivative is
\[-3\sin(x)+3\cos^2(x)\sin(x)\] right?

Answer 56

-3sin^3 x is the first derivative

Answer 57

so it is

Answer 58

and therefore your function is increasing on \((\pi, 2\pi)\) and decreasing on \((0,\pi)\)

Answer 59

as the cube doesn't change the sign, but the -3 out front does. so far so good?

Answer 60

so far so good

Answer 61

second derivative
\[f''(x)=-9\sin^2(x)\cos(x)\]

Answer 62

You are so intelligent @satellite73

Answer 63

now we only need to worry about
\[\cos(x)\] because \(\sin^2(x)\geq 0\)
on the interval \((\frac{\pi}{2},\frac{3\pi}{2})\) we have \(\cos(x)<0\)and so \(-\cos(x)>0\) and that is where your original funcion is concave up

Answer 64

it will be concave down on \((0,\frac{\pi}{2})\) and also on \((\frac{3\pi}{2},2\pi)\)

Answer 65

yea that works thanks

Answer 66

that about takes care of it. not sure what "works" means, but i guess it is on line homework

Answer 67

its an online hw yea...do you know the point of inflection