Question

[Nguyễn Ngọc Ánh]

Tìm GTLN,GTNN của hàm số

y=4/3*(sin⁶x+cos⁶x)+cos4x-1

Answer 1

[Nguyễn Hoàng Minh]

\(y=\dfrac{4}{3}\left(\sin^6x+\cos^6x\right)+\cos4x-1\)

\(\sin^6x+\cos^6x=\left(\sin^2x+\cos^2x\right)\left(\sin^4x-\sin^2x\cdot\cos^2x+\cos^4x\right)\\ =\left(\sin^2x+\cos^2x\right)^2-3\sin^2x\cdot\cos^2x=1-\dfrac{3}{4}\sin^22x\)

Do \(0\le\sin^22x\le1\Leftrightarrow\dfrac{3}{4}\cdot0\ge-\dfrac{3}{4}\sin^22x\ge-\dfrac{3}{4}\)

\(\Leftrightarrow1\ge1-\dfrac{3}{4}\sin^22x\ge1-\dfrac{3}{4}=\dfrac{1}{4}\\ \Leftrightarrow\dfrac{4}{3}\ge\dfrac{4}{3}\left(\sin^6x+\cos^6x\right)\ge\dfrac{1}{4}\cdot\dfrac{4}{3}=\dfrac{1}{3}\)

Ta có \(-1\le\cos4x\le1\)

\(\Leftrightarrow\dfrac{1}{3}-1-1\le\dfrac{4}{3}\left(\sin^6x+\cos^6x\right)+\cos4x-1\le\dfrac{4}{3}+1-1\\ \Leftrightarrow-\dfrac{5}{3}\le y\le\dfrac{4}{3}\)

Vậy \(y_{min}=-\dfrac{5}{3};y_{max}=\dfrac{4}{3}\)

Answer 2

[Nguyễn Việt Lâm]

\(y=\dfrac{4}{3}\left(sin^6x+cos^6x\right)+cos4x-1\)

\(y=\dfrac{4}{3}\left(\dfrac{5}{8}+\dfrac{3}{8}cos4x\right)+cos4x-1\)

\(y=\dfrac{3}{2}cos4x-\dfrac{1}{6}\)

\(-1\le cos4x\le1\Rightarrow-\dfrac{5}{3}\le y\le\dfrac{4}{3}\)

\(y_{min}=-\dfrac{5}{3}\) khi \(cos4x=-1\)

\(y_{max}=\dfrac{4}{3}\) khi \(cos4x=1\)