Question

Cho \(y=x^3-3x^2+2\). Tìm x để :

a) \(y'>0\)

b) \(y'< 3\)

Answer 1

\(y'=x^2-6x\)

y' > 0 =>x<0;6<x

y' <3=>\(3-2\sqrt{3}< x< 3+2\sqrt{3}\)

Answer 2

\(y'\left(x\right)=3x^2-6x\).
a) \(y'\left(x\right)>0\)\(\Leftrightarrow3x^2-6x>0\)\(\Leftrightarrow\left[{}\begin{matrix}x< 0\\x>2\end{matrix}\right.\).
Vậy \(x< 0\) hoặc \(x>2\) thì \(y'\left(x\right)>0\).
b) \(y'\left(x\right)< 3\)\(\Leftrightarrow3x^2-6x< 3\)\(\Leftrightarrow3x^2-6x-3< 0\)\(\Leftrightarrow1-\sqrt{2}< x< 1+\sqrt{2}\).
Vậy \(1-\sqrt{2}< x< 1+\sqrt{2}\) thì \(y'\left(x\right)< 3\).