Question

[Nguyễn Khánh Linh]

Tìm gtln và gtnn ạ. Mình cần xem cách giải ạ

Answer 1

[Nguyễn Việt Lâm]

a.

\(f'\left(x\right)=2cos2x-1=0\Rightarrow cos2x=\dfrac{1}{2}\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}\\x=-\dfrac{\pi}{6}\end{matrix}\right.\)

Ta có:

\(f\left(-\dfrac{\pi}{2}\right)=0+\dfrac{\pi}{2}=\dfrac{\pi}{2}\)

\(f\left(\dfrac{\pi}{2}\right)=0-\dfrac{\pi}{2}=-\dfrac{\pi}{2}\)

\(f\left(\dfrac{\pi}{6}\right)=\dfrac{\sqrt{3}}{2}-\dfrac{\pi}{6}\)

\(f\left(-\dfrac{\pi}{6}\right)=-\dfrac{\sqrt{3}}{2}+\dfrac{\pi}{6}\)

So sánh các giá trị trên ta được:

\(f\left(x\right)_{max}=f\left(-\dfrac{\pi}{2}\right)=\dfrac{\pi}{2}\)

\(f\left(x\right)_{min}=f\left(\dfrac{\pi}{2}\right)=-\dfrac{\pi}{2}\)

Answer 2

[Nguyễn Việt Lâm]

b.

\(f'\left(x\right)=3-2\sqrt{3}sin2x=0\Rightarrow sin2x=\dfrac{\sqrt{3}}{2}\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}\\x=\dfrac{\pi}{3}\end{matrix}\right.\)

Ta có: \(f\left(-\dfrac{\pi}{2}\right)=-\dfrac{3\pi}{2}-\sqrt{3}\)

\(f\left(\dfrac{\pi}{6}\right)=\dfrac{\pi}{2}+\dfrac{\sqrt{3}}{2}\)

\(f\left(\dfrac{\pi}{3}\right)=\pi-\dfrac{\sqrt{3}}{2}\)

\(f\left(\pi\right)=3\pi+\sqrt{3}\)

Từ đó: \(f_{min}=f\left(-\dfrac{\pi}{2}\right)=-\dfrac{3\pi}{2}-\sqrt{3}\)

\(f_{max}=f\left(\pi\right)=3\pi+\sqrt{3}\)