Chứng minh các biểu thức đã cho không phụ thuộc vào x.

Từ đó suy ra f'(x)=0

a) ;

b) ;

c) (\(\sqrt{2}\)-\(\sqrt{6}\))=>f'(x)=0

d,f(x)=\(\frac{3}{2}\)=>f'(x)=0

\(f\left(-x\right)=\left|-sinx-cosx\right|-\left|-sinx+cosx\right|\)

\(=\left|sinx+cosx\right|-\left|sinx-cosx\right|=-f\left(x\right)\)

\(\Rightarrow f\left(x\right)+f\left(-x\right)=0\)

\(\Rightarrow T=f\left(-\pi\right)+f\left(\pi\right)+f\left(-\frac{\pi}{2}\right)+f\left(\frac{\pi}{2}\right)+...+f\left(-\frac{\pi}{n}\right)+f\left(\frac{\pi}{n}\right)+f\left(0\right)\)

\(=0+0+...+0+f\left(0\right)=f\left(0\right)\)

\(=1-1=0\)

Hàm số xác định khi: \(\left\{{}\begin{matrix}tanx\ne\pm1;cosx\ne0\\cosx\ne-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm\dfrac{\pi}{4}+k\pi\\x\ne\dfrac{\pi}{2}+k\pi\\x\ne\pi+k2\pi\end{matrix}\right.\)

Ta có \(f'\left( x \right) = 2.2\sin \left( {x + \frac{\pi }{4}} \right).{\left[ {\sin \left( {x + \frac{\pi }{4}} \right)} \right]^,} = 4\sin \left( {x + \frac{\pi }{4}} \right)\cos \left( {x + \frac{\pi }{4}} \right) = 2\sin \left( {2x + \frac{\pi }{2}} \right)\)

\( \Rightarrow f''\left( x \right) = 2.2\cos \left( {2x + \frac{\pi }{2}} \right) = 4\cos \left( {2x + \frac{\pi }{2}} \right)\)

Mặt khác \( - 1 \le \cos \left( {2x + \frac{\pi }{2}} \right) \le 1 \Leftrightarrow  - 4 \le f''\left( x \right) \le 4\)

Vậy \(\left| {f''\left( x \right)} \right| \le 4\) với mọi x.

a/ \(f'\left(x\right)=12sin^33x.cos3x\)

\(f'\left(x\right)=g\left(x\right)\Leftrightarrow12sin^33x.cos3x=sin6x\)

\(\Leftrightarrow6sin^23x.2sin3x.cos3x-sin6x=0\)

\(\Leftrightarrow6sin^23x.sin6x-sin6x=0\)

\(\Leftrightarrow sin6x\left(6sin^23x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}sin6x=0\\sin^23x=\frac{1}{6}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}sin6x=0\\\frac{1-cos6x}{2}=\frac{1}{6}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin6x=0\\cos6x=\frac{2}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}6x=k\pi\\6x=a+k2\pi\\6x=-a+k2\pi\end{matrix}\right.\) với \(cosa=\frac{2}{3}\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{k\pi}{6}\\x=\frac{a}{6}+\frac{k\pi}{3}\\x=-\frac{a}{6}+\frac{k\pi}{3}\end{matrix}\right.\)

b/

\(f'\left(x\right)=6sin^22x.cos2x=4cos2x-5sin4x\)

\(\Leftrightarrow6sin^22x.cos2x=4cos2x-10sin2x.cos2x\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\Rightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\\3sin^22x=2-5sin2x\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow3sin^22x+5sin2x-2=0\)

\(\Rightarrow\left[{}\begin{matrix}sin2x=\frac{1}{3}\\sin2x=-2< -1\left(l\right)\end{matrix}\right.\)

\(\Rightarrow sin2x=sina\) (với \(sina=\frac{1}{3}\))

\(\Rightarrow\left[{}\begin{matrix}2x=a+k2\pi\\2x=\pi-a+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{a}{2}+k\pi\\x=\frac{\pi}{2}-\frac{a}{2}+k\pi\end{matrix}\right.\)

c/

\(f'\left(x\right)=4x.cos^2\frac{x}{2}-2x^2.cos\frac{x}{2}.sin\frac{x}{2}=2x\left(1+cosx\right)-x^2sinx\)

\(f'\left(x\right)=g\left(x\right)\)

\(\Leftrightarrow2x\left(1+cosx\right)-x^2sinx=x-x^2sinx\)

\(\Leftrightarrow2x\left(1+cosx\right)=x\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2\left(1+cosx\right)=1\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow cosx=-\frac{1}{2}\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{2\pi}{3}+k2\pi\\x=-\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)

\(f'\left(x\right)=4sin\left(3x-\dfrac{\pi}{4}\right)\cdot\left[sin\left(3x-\dfrac{\pi}{4}\right)\right]'\\ =4\left(3x-\dfrac{\pi}{4}\right)'cos\left(3x-\dfrac{\pi}{4}\right)sin\left(3x-\dfrac{\pi}{4}\right)\\ =6sin\left(6x-\dfrac{\pi}{2}\right)\)

Vì \(-1\le sin\left(6x-\dfrac{\pi}{2}\right)\le1\Rightarrow-6\le6sin\left(6x-\dfrac{\pi}{2}\right)\le6\Leftrightarrow-6\le f'\left(x\right)\le6\)

Vậy \(\left|f'\left(x\right)\right|\le6\forall x\)