Chứng minh rằng \(f'\left( x \right) > 0,\forall x \in R\), nếu
a) Ta có:
\(\begin{array}{l}
f\prime \left( x \right) = 6{x^8} - 6{x^5} + 6{x^2} - 6x + 6\\
= 6\left( {{x^8} - {x^5} + {x^2} - x + 1} \right)\\
= 6{x^2}\left( {{x^6} - {x^3} + \frac{1}{4}} \right) + 3{x^2} + 6\left( {\frac{{{x^2}}}{4} - x + 1} \right)\\
= 6{x^2}{\left( {{x^3} - \frac{1}{2}} \right)^2} + 3{x^2} + 6\left( {\frac{x}{2} - 1} \right)2 > 0,\forall x \in R
\end{array}\)
b) \(f'\left( x \right) = 2 + \cos x\) vì \(|\cos x|\, \le 1 \Rightarrow 2 + \cos x > 0,\forall x \in R\)
-- Mod Toán 11