y= căn 3x2 + x3 tính đơn điệu

Đáp án:

$\eqalign{ & Ham\,\,so\,\,DB/\,\,\left( { - 3; - 2} \right);\,\,\left( {0; + \infty } \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,NB/\left( { - 2;0} \right) \cr}$

Giải thích các bước giải:

$\eqalign{ & y = \sqrt {3{x^2} + {x^3}} \cr & DKXD:\,\,3{x^2} + {x^3} \ge 0 \Leftrightarrow {x^2}\left( {x + 3} \right) \ge 0 \Leftrightarrow x \ge - 3 \cr & \Rightarrow D = \left[ { - 3; + \infty } \right) \cr & Ta\,\,co:\,\,y' = {{6x + 3{x^2}} \over {2\sqrt {3{x^2} + {x^3}} }} \cr & Cho\,\,y' = 0 \Leftrightarrow 6x + 3{x^2} = 0 \Leftrightarrow \left[ \matrix{ x = 0 \hfill \cr x = - 2 \hfill \cr} \right.\,\,\left( {tm} \right) \cr & BXD:\, \cr & - 3\,\,\,\,\,\,\, + \,\,\,\,\,\,\,\,\,\, - 2\,\,\,\,\,\,\,\,\,\,\, - \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \infty \cr & \Rightarrow Ham\,\,so\,\,DB/\,\,\left( { - 3; - 2} \right);\,\,\left( {0; + \infty } \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,NB/\left( { - 2;0} \right) \cr}$