How can I find critical number of f(x)=x/(x^2-x+1) [0,3]. When I solve 3x^2-2x+1 , it is complex number, so, I am confused about the result.

Question

How can I find critical number of  f(x)=x/(x^2-x+1) [0,3].  When I solve 3x^2-2x+1 , it is complex number, so, I am confused about the result.

anonymous · Answer

to find the critical cnumber, you first obtain its derivative

anonymous · Answer

yes, when I solve the derivative , it is not real numbers

anonymous · Answer

$$
f(x)=\frac{x}{x^2-x+1}\
f'(x)=\frac{1(x^2-x+1)-(x)(2x-1)}{(x^2-x+1)}=0\
\implies x^2-x+1-2x^2+x=0\
\implies -x^2+1=0\implies x^2-1=0\
\implies (x-1)(x+1)=0
$$

anonymous · Answer

@Yating  can you take it from there?

anonymous · Answer

Yes, thanks, I did differently from what you showed. Let me check where I made mistakes

anonymous · Answer

Oh, I misused quotient rule. Thanks a lot.

anonymous · Answer

yep

anonymous · Answer

also, I missed to put the square part in the denominator of the quotient rule

anonymous · Answer

I figured out, but the most important part in this case is numerator.