Chứng minh rằng \(f'(x) = 0,\forall x \in R\), nếu:
a)
f\left( x \right) = 3{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} - 6{\sin ^2}x{\cos ^2}x\\
- 2\left( {{{\sin }^4}x - {{\sin }^2}x{{\cos }^2}x + {{\cos }^4}x} \right)\\
= 3 - 6{\sin ^2}x{\cos ^2}x - 2\left[ {\left( {{{\sin }^2}x + {{\cos }^2}x} \right) - 3{{\sin }^2}x{{\cos }^2}x} \right]\\
= 3 - 6{\sin ^2}x{\cos ^2}x - 2 + 6{\sin ^2}x{\cos ^2}x = 1 \Rightarrow f\prime \left( x \right) = 0
\end{array}\)
b)
\(\begin{array}{l}
f(x) = ({\cos ^6}x + {\sin ^2}x{\cos ^4}x) + (2{\sin ^2}x{\cos ^4}x + 2{\sin ^4}x{\cos ^2}x) + {\sin ^4}x\\
= {\cos ^4}x + 2{\sin ^2}x{\cos ^2}x + {\sin ^4}x\\
= {({\sin ^2}x + {\cos ^2}x)^2} = 1 \Rightarrow f\prime (x) = 0
\end{array}\)
c)
\(\begin{array}{l}
f\left( x \right) = \cos \left( {x - \frac{\pi }{3}} \right)\cos \left( {x + \frac{\pi }{4}} \right) + \sin \left( {\frac{\pi }{3} - x} \right)\sin \left( { - \frac{\pi }{4} - x} \right)\\
= \cos \left( {x - \frac{\pi }{3}} \right)\cos \left( {x + \frac{\pi }{4}} \right) + \sin \left( {x - \frac{\pi }{3}} \right)\sin \left( {x + \frac{\pi }{4}} \right)\\
= \cos \left[ {\left( {x - \frac{\pi }{3}} \right) - \left( {x + \frac{\pi }{4}} \right)} \right] = \cos \frac{{7\pi }}{{12}} \Rightarrow f\prime \left( x \right) = 0
\end{array}\)
d)
\(\begin{array}{l}
f(x) = \frac{1}{2}\left[ {1 + \cos 2x + 1 + \cos \left( {\frac{{4\pi }}{3} + 2x} \right) + 1 + \cos \left( {\frac{{4\pi }}{3} - 2x} \right)} \right]\\
= \frac{1}{2}\left[ {3 + \cos 2x + 2.\cos \frac{{4\pi }}{3}\cos 2x} \right] = \frac{3}{2} \Rightarrow f\prime (x) = 0
\end{array}\)
-- Mod Toán 11