Cho $f\left( x \right)$ là hàm bậc bốn thỏa mãn $f\left( 0...

Hàm số là hàm bậc ba, đạt cực trị tại các điểm và nên ta có:

Từ bảng biến thiên ta có: $\left\{ \begin{aligned}
& {f}'\left( -5 \right)= -\dfrac{53}{3} \\
& {f}'\left( -3 \right)= -19 \\
\end{aligned} \right.$$\Rightarrow \left\{ \begin{aligned}
& a=1 \\
& d=-1 \\
\end{aligned} \right.\Rightarrow {f}'\left( x \right)= \dfrac{1}{3}{{x}^{3}}+ 4{{x}^{2}}+15x-1= \dfrac{1}{3}x\left( {{x}^{2}}+ 4x+15 \right)-1h\left( x \right)=f\left( {{x}^{5}} \right)-x-2{h}'\left( x \right)=5{{x}^{4}}{f}'\left( {{x}^{5}} \right)-1{h}'\left( x \right)=0\Leftrightarrow {f}'\left( {{x}^{5}} \right)=\dfrac{1}{5{{x}^{4}}}\left( 1 \right)x=0\bullet x<0{f}'\left( x \right)<0{f}'\left( {{x}^{5}} \right)<0 \forall x<0\Rightarrow \left( 1 \right)\left( -\infty ; 0 \right)\bullet x>0\Rightarrow {f}'\left( x \right)\left( 0 ; +\infty \right)\Rightarrow {f}'\left( {{x}^{5}} \right)\left( 0 ; +\infty \right)y=\dfrac{1}{5{{x}^{4}}}\left( 0 ; +\infty \right)\Rightarrow \left( 1 \right)\left( 0 ; +\infty \right)\underset{x\to {{0}^{+}}}{\mathop{\lim }} \left[ {f}'\left( {{x}^{5}} \right)-\dfrac{1}{5{{x}^{4}}} \right]=-\infty \underset{x\to +\infty }{\mathop{\lim }} \left[ {f}'\left( {{x}^{5}} \right)-\dfrac{1}{5{{x}^{4}}} \right]=+\infty u\left( x \right)={f}'\left( {{x}^{5}} \right)-\dfrac{1}{5{{x}^{4}}}\left( 0 ; +\infty \right)\Rightarrow \left( 1 \right)\left( 0 ; +\infty \right)\left( 1 \right){{x}_{0}}\in \left( 0 ; +\infty \right)h\left( x \right)=f\left( {{x}^{5}} \right)-x-2 Do \)">h\left( 0 \right)=-2<0h\left( {{x}_{0}} \right)<0\Rightarrow h\left( x \right)=0\Rightarrow g\left( x \right)=\left| h\left( x \right) \right|=\left| f\left( {{x}^{5}} \right)-x-2 \right|3$ điểm cực trị.