Cho hàm số bậc ba \(y = a{x^3} + b{x^2} + cx + d\) có đồ thị như hình vẽ bên.

* Hướng dẫn giải

\(f\left( 1 \right) = 2,f\left( {{x_0}} \right) = f\left( 2 \right) = 0,f\left( {{x_1}} \right) = f\left( {{x_2}} \right) = f\left( {{x_3}} \right) = 1\) 

Xét hàm số \(g\left( x \right) = \frac{{\left( {{x^2} - 4x + 4} \right)\sqrt {x - 1} }}{{x\left[ {{f^2}\left( x \right) - f\left( x \right)} \right]}}\) có TXĐ: \(\left\{ \begin{array}{l}
x \ge 1\\
x \ne 0\\
f\left( x \right) \ne 0\\
f\left( x \right) \ne 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
x \ne 0\\
x \ne {x_0}\\
x \ne {x_1}\\
x \ne {x_2}\\
x \ne {x_3}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
x \ne {x_2}\\
x \ne {x_3}
\end{array} \right.,1 < {x_2} < 2 < {x_3}\) 

\(\mathop {\lim }\limits_{x \to {x_2}} g\left( x \right) = \mathop {\lim }\limits_{x \to {x_2}} \frac{{\left( {{x^2} - 4x + 4} \right)\sqrt {x - 1} }}{{x\left[ {{f^2}\left( x \right) - f\left( x \right)} \right]}} = \infty ;\mathop {\lim }\limits_{x \to {x_3}} g\left( x \right) = \mathop {\lim }\limits_{x \to {x_3}} \frac{{\left( {{x^2} - 4x + 4} \right)\sqrt {x - 1} }}{{x\left[ {{f^2}\left( x \right) - f\left( x \right)} \right]}} = \infty \) 

Suy ra đths \(g\left( x \right) = \frac{{\left( {{x^2} - 4x + 4} \right)\sqrt {x - 1} }}{{x\left[ {{f^2}\left( x \right) - f\left( x \right)} \right]}}\) có 2 đường tiệm cận đứng.