Can someone please help me!!!!!

[Guest]

I am taking cal.1 rigth now and i have been trying all this moring to solve this problem. It is a graphing problem. I need to find the inflection pt @ x=(1/4)^1/3 or cube root of 1/4
so i know i have to to plug this into the orginal funtion and solve for x, but i don't now how to SO CAN SOMEONE PLEASE HELP ME!! Here is the problem

g(x)=2X^5 -- 2X^2
g(cube root of 1/4 or (1/4)^1/3)= 2((1/4)^1/3)^5 -- 5((1/4)^1/3)^2

ACCORDING TO THE SOULTION IN THE BACK OF THE BOOK THE ANSWER IS 1-9/2((4)^1/3)^2, i have not clue how they got that ans.
SO SOMEONE PLEASE HELP : :shock: [/list][/code]

 

[pka]

Please post the exact wording of the question.

 

[Guest]

Say it again? I don't understand what you are trying to ask. Point of inflection is second derivative....

So if you have
g(x)=2x^5-2x^2

g'(x)= 10x^4-4x

g''(x)= 30x^3-4

0=30x^3-4
4/30=x^3

(2/15)^1/3 = x

The point of inflection (I don't know what you are trying to ask) is at 0.5108 or x= cube root of 2/15

or scan the problem haha ... it'll be nice to see what you are trying to ask.

 

[soroban]

Hello, cal.1brainhurt!

I need to find the inflection pt @ x = (1/4)^1/3

so i know i have to to plug this into the orginal function and solve for x . . . what?
but i don't now how to . . . neither do I ... we already know the value of x, don't we?



SO CAN SOMEONE PLEASE HELP ME!! Here is the problem

g(x)=2X⁵ - 2X² . . . obviously a typo here!

g([1/4]^1/3) .= .2([1/4]^1/3)⁵ - 5([1/4]^1/3)² . simplify this carefully


ACCORDING TO THE SOULTION IN THE BACK OF THE BOOK
THE ANSWER IS 1 - 9/2((4)^1/3)^2 . . . I don't think so!

 

[Guest]

Hello, cal.1brainhurt!

I need to find the inflection pt @ x = (1/4)^1/3

so i know i have to to plug this into the orginal function and solve for x . . . what?
but i don't now how to . . . neither do I ... we already know the value of x, don't we?



SO CAN SOMEONE PLEASE HELP ME!! Here is the problem

g(x)=2X⁵ - 2X² . . . obviously a typo here!

g([1/4]^1/3) .= .2([1/4]^1/3)⁵ - 5([1/4]^1/3)² . simplify this carefully


ACCORDING TO THE SOULTION IN THE BACK OF THE BOOK
THE ANSWER IS 1 - 9/2((4)^1/3)^2 . . . I don't think so!

i am sorry i made a typo :
g(x)=2X^5-5X^2
g([1/4]^1/3)= 2([1/4]^1/3)^5--5([1/4]^1/3)^2

SORRY ABOUT THAT
PLEASE CAN YOU SHOW ME HOW TO SOLVE THIS AND GET THE ANSWER
1-9/2[4^1/3]^2

 

[Guest]

CAN SOMEONE HELP!!

i am sorry i made a TYPO :
g(x)=2X^5-5X^2
g([1/4]^1/3)= 2([1/4]^1/3)^5--5([1/4]^1/3)^2

SORRY ABOUT THAT
PLEASE CAN YOU SHOW ME HOW TO SOLVE THIS AND GET THE ANSWER
1-9/2[4^1/3]^2

 

[tkhunny]

Re: CAN SOMEONE HELP!!

g(x)=2X^5-5X^2
g([1/4]^1/3)= 2([1/4]^1/3)^5--5([1/4]^1/3)^2

So, you are saying you already know the x-coordinate of the point of inflection and you seek the y-coordinate, which is a matter of substituting in to the function. You've already "solved". You need to "simplify".

I have a plan that can simplify your life. BEFORE substituting, rewrite g(x) into a more conveniant form.

g(x) = 2x⁵-5x² = (x²)*(2*x³ - 5)

Something tells me that x³ is going to come in handy.

Note: I still don't know what that "answer" is. The only way I can see to get that answer is to multiply the correct answer by zero (0) and add what is wanted. It's a little foolish, but, hey, it works.

 

[Guest]

THANK YOU SOOOOOOOOOOOOOOOOOOOO MUCH=)=)

YOUR PALN WORKED!! i had been working on this problem for two days. I could no figure out how the book got 1-9/2[4^1/3)^2 as the answer, but thanks to you i figured it out. I can't thank you enought. You were sent my God to save me. THANK YOU!!! THANK YOU!!THANKYOU!! THANK YOU!! THANK YOU!!! THANK YOU!! SOOOOOOOOOOOOOOOOOOO MUCH. :lol:

 

[tkhunny]

Ummm....You managed that 1 - 9/2 thing? How?

 

[Guest]

Huh, heh?

 

[soroban]

Re: THANK YOU SOOOOOOOOOOOOOOOOOOOO MUCH=)=)

Hello, all!

I have a theory . . .
[I'm the head of MSI, Math Scene Investiations, you see.]

The book got 1-9/2[4^1/3)^2 as the answer.

First of all, I seriously doubt that the book printed the answer that way.

What author would write (4^1/3)² instead of 4^2/3 ?
. . and wouldn't that be simplified to: .2·2^1/3 ?

All this makes me suspicious of that leading "1" . . . I bet it's <u>multiplication</u>!

Plugging in x = (1/4)^1/3
. . . . . . . . . . . . . . . . . . . 9
. . the answer is: . - ------------
. . . . . . . . . . . . . . . . 2(1/4)^2/3

. . which has an amazing resemblence to the so-called book-answer.