Andrew Rubin
New member
- Joined
- Jun 24, 2019
- Messages
- 22
Hi everyone
I'm working on an optimalization problem, although I'm stuck on an algebra simplication issue.
First, I obtain the derivative of
[MATH]f\left(x\right)\:=\:x^3-3x^2+3x[/MATH]as,
[MATH]f'\left(x\right)\:=\:3x^2-6x+3[/MATH]
Then my textbook says that this can be further simplified to,
[MATH]f'\left(x\right)\:=\:3\left(x-1\right)^2[/MATH]
I can't crack how to arrive at this simplication. I've tried to first cancel out the common factor 3 (don't know if this is correct terminology for my use here) from
[MATH]f'\left(x\right)\:=\:3x^2-6x+3[/MATH]to arrive at,
[MATH]f'\left(x\right)\:=\:3\left(x^2-6x\right)[/MATH]
Then I progress using the law of distribution of multiplication, [MATH]a(b±c)≡ab±ac[/MATH], and arrive at
[MATH]=\:3\cdot x^2-3\cdot 6x[/MATH]
But I'm not sure if this is the right path nor how to move forward. I would greatly appreciate input on this.
I'm working on an optimalization problem, although I'm stuck on an algebra simplication issue.
First, I obtain the derivative of
[MATH]f\left(x\right)\:=\:x^3-3x^2+3x[/MATH]as,
[MATH]f'\left(x\right)\:=\:3x^2-6x+3[/MATH]
Then my textbook says that this can be further simplified to,
[MATH]f'\left(x\right)\:=\:3\left(x-1\right)^2[/MATH]
I can't crack how to arrive at this simplication. I've tried to first cancel out the common factor 3 (don't know if this is correct terminology for my use here) from
[MATH]f'\left(x\right)\:=\:3x^2-6x+3[/MATH]to arrive at,
[MATH]f'\left(x\right)\:=\:3\left(x^2-6x\right)[/MATH]
Then I progress using the law of distribution of multiplication, [MATH]a(b±c)≡ab±ac[/MATH], and arrive at
[MATH]=\:3\cdot x^2-3\cdot 6x[/MATH]
But I'm not sure if this is the right path nor how to move forward. I would greatly appreciate input on this.