Đáp án:
\(min_{y}=-\dfrac{19}{4}\)
\(max_{y}=-4\)
Giải thích các bước giải:
\(y=\sin^{6} x+\cos^{6} -5\)
\(\Leftrightarrow y=(\sin^{2} x)^{3}+(\cos^{2} x)^{3}-5\)
\(\Leftrightarrow y=(\sin^{2} x+\cos^{2} x)(\sin^{4} x-\sin^{2} x.\cos^{2} x+\cos^{4} x)-5\)
\(\Leftrightarrow y=(\sin^{2} x+\cos^{2} x)^{2}-3\sin^{2} x.\cos^{2} x-5\)
\(\Leftrightarrow y=1-\dfrac{3}{4}\sin^{2} 2x-5\)
\(\Leftrightarrow y=-4-\dfrac{3}{4} \sin^{2} 2x\)
Ta có: \(0 \leq \sin^{2} 2x \leq 1\)
\(\Leftrightarrow -\dfrac{3}{4} \leq -\dfrac{3}{4} \sin^{2} 2x \leq 0\)
\(\Leftrightarrow \dfrac{-19}{4} \leq -4-\dfrac{3}{4} \sin^{2} 2x \leq -4\)
Vậy \(min_{y}=-\dfrac{19}{4}\)
\(max_{y}=-4\)
