Recent content by Orson

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.

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

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

Can you help me see where it's wrong ?

(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

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

Are you saying get rid of the -2

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 ?

I'll start again.

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

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

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

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...

It was multiple choice on khan academy. for which values of x is the function a local minimum or maximum. i can't remember which. But -1/3 was the correct answer per the software (s-word again)