Find inflection points of polynomials.

[Orson]

Homework Statement

Find critical points and inflection points of:
1/[x(x-1)]

Homework Equations

1/[x(x-1)]

The Attempt at a Solution

using quotient rule, we obtain
(0-(2x-1)/(x^2-x)^2
set -2x+1=0 we get 1/2 for critical point.

for second derivative,
i get -2(x^2-x)^2-(-2x+1)4x(x^2-x)(2x-1)
the -(-2x+1) gives two terms of (2x-1) but i have no idea how to factor out both the (x^2-x) and the( 2x-1) and have only terms remaining that are joined by multiplication.

 

[mfb]

Factor out (x^2-x), and then just multiply out everything else and see what you get. You'll see another factor to take out later.

 

[Orson]

Factor out (x^2-x), and then just multiply out everything else and see what you get. You'll see another factor to take out later.

i don't know what to do with the extra (2x-1)

 

[Orson]

i don't know what to do with the extra (2x-1)

also that -2, can that move in front of the derivative?

 

[Orson]

I get left with
-2(x^2-x)(x^2-x)+2x-1(4x+1)

 

[mfb]

i don't know what to do with the extra (2x-1)

Multiply out everything.

-2(x^2-x)(x^2-x)+2x-1(4x+1)

It looks like you made more than one error. It would help to see the steps.

 

[Orson]

Multiply out everything.It looks like you made more than one error. It would help to see the steps.

I'll start again.

 

[Orson]

Ok first step in factoring out (x^2-x) I get

-2(x^2-x)(x^2-x)+(2x-1)(4x)(2x-1)

Should I have canceled the second (x^2-x) because of the denominator ?

 

[mfb]

-2(x^2-x)(x^2-x)+(2x-1)(4x)(2x-1)

That doesn't have (x²-x) factored out. Otherwise it would look like (x²-x)(something)

 

[Orson]

That doesn't have (x²-x) factored out. Otherwise it would look like (x²-x)(something)

Are you saying get rid of the -2

 

[Orson]

Ok first step in factoring out (x^2-x) I get

-2(x^2-x)(x^2-x)+(2x-1)(4x)(2x-1)

Should I have canceled the second (x^2-x) because of the denominator ?

 

[Orson]

(x^2-x)(x^2-x)+(2x-1)(4x)(2x-1)(-2)

 

[Orson]

(x^2+x)(x^2+x) +(2x-1)(4x)(2x-1)(-2)
= (x^2+x)(x^2+x)+ -8x(4x^2-4x+1)
=(x^2+x)(x^2+x)+ -32x^3-32x^2-8x

 

[mfb]

It should look like (x²-x)(something). In other words, (x²-x) multiplied by something is the full expression. As long as it doesn't look like that, you didn't factor out (x²-x) properly.

 

[Orson]

It should look like (x²-x)(something). In other words, (x²-x) multiplied by something is the full expression. As long as it doesn't look like that, you didn't factor out (x²-x) properly.

Can you help me see where it's wrong ?

 

[Orson]

Can you help me see where it's wrong ?

Should the parenthesis come off the second factor ? So the values are included with the rest ?

 

[mfb]

Can you help me see where it's wrong ?

I don't understand what is unclear.

$-2(x^2-x)^2-(-2x+1)4x(x^2-x)(2x-1) = (x^2-x)(something)$ - find "something"

 

[Orson]

I don't understand what is unclear.

$-2(x^2-x)^2-(-2x+1)4x(x^2-x)(2x-1) = (x^2-x)(something)$ - find "something"

I am having trouble finding the something. Is that clear?

 

[LCKurtz]

Perhaps you need to review your algebra. You are having the same problem factoring that we discussed in your earlier post. You are factoring out $(x^2 - x)$ and the "something" that goes in the other parentheses is what is left. You did it before after some prompting and hopefully you can do it again.

 

[Orson]

Perhaps you need to review your algebra. You are having the same problem factoring that we discussed in your earlier post. You are factoring out $(x^2 - x)$ and the "something" that goes in the other parentheses is what is left. You did it before after some prompting and hopefully you can do it again.

i am quite aware that the (x^2-x) comes out. I have been for the last 12 hours. I am having trouble with what comes after it. But I'll go review my algebra. Thank you both for nothing.