Question

Cho \(f\left(x\right)=\left(x-2\right)\left(\sqrt{3x^2+1}\right)\). giải pt f(x)' \(^2\) =0?

Answer 1

Đề là \(f''\left(x\right)=0\) hay \(\left[f'\left(x\right)\right]^2=0\) nhỉ?

Answer 2

\(f'\left(x\right)=\sqrt{3x^2+1}+\dfrac{3x\left(x-2\right)}{\sqrt{3x^2+1}}=\dfrac{6x^2-6x+1}{\sqrt{3x^2+1}}\)

\(\left[f'\left(x\right)\right]^2=0\Leftrightarrow f'\left(x\right)=0\Leftrightarrow6x^2-6x+1=0\)

\(\Rightarrow x=\dfrac{3\pm\sqrt{3}}{6}\)