Cho $f(x)$ là hàm số bậc bốn thỏa mãn $f(0)=0.$ Hàm số...

Xét hàm số

$$(Chỉ xét $x\ne 0x=0{{x}^{3}}=u\Rightarrow {{x}^{2}}=\sqrt[3]{{{u}^{2}}}\left( * \right)f'\left( u \right)=\frac{2021}{3\sqrt[3]{{{u}^{2}}}}\left( * \right)y=f'\left( u \right)y=\frac{2021}{3\sqrt[3]{{{\mathsf{u}}^{2}}}}y=t\left( u \right)=\frac{2021}{3\sqrt[3]{{{u}^{2}}}}\Rightarrow t'\left( u \right)=-\frac{4042}{9}.\frac{1}{\sqrt[3]{{{u}^{5}}}} \)">\Rightarrow y=f'\left( u \right)y=\frac{2021}{3\sqrt[3]{{{u}^{2}}}} Dựa vào ĐTHS, ta thấy đồ thị hàm \)">y=f'\left( u \right)y=\frac{2021}{3\sqrt[3]{{{u}^{2}}}}a\Rightarrow f'\left( u \right)=\frac{2021}{3\sqrt[3]{{{u}^{2}}}}u=a>0\Rightarrow \left( * \right)1x=\sqrt[3]{a}\Rightarrow h'\left( x \right)=01x=\sqrt[3]{a}h\left( x \right) (Giải thích \)">\left( 1 \right)h\left( 0 \right)=f\left( 0 \right)-0=0y=h\left( x \right)y=g\left( x \right)=\left| h\left( x \right) \right| \)">\Rightarrow g\left( x \right)3$ cực trị