Toán Lớp 11: Tìm GTLLN, GTNN của:
y = 2 cos^2x – cosx – 3
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Toán Lớp 11: Tìm GTLLN, GTNN của:
y = 2 cos^2x – cosx – 3
Comments ( 1 )
y = 2{\cos ^2}x – \cos x – 3\\
= 2.\left( {{{\cos }^2}x – \dfrac{1}{2}\cos x} \right) – 3\\
= 2.\left( {{{\cos }^2}x – 2.\cos x.\dfrac{1}{4} + \dfrac{1}{{16}}} \right) – 2.\dfrac{1}{{16}} – 3\\
= 2.{\left( {\cos x – \dfrac{1}{4}} \right)^2} – \dfrac{{25}}{8}\\
Do: – 1 \le \cos x \le 1\\
\Leftrightarrow – \dfrac{5}{4} \le \cos x – \dfrac{1}{4} \le \dfrac{3}{4}\\
\Leftrightarrow 0 \le {\left( {\cos x – \dfrac{1}{4}} \right)^2} \le \dfrac{{25}}{{16}}\\
\Leftrightarrow 0 \le 2.{\left( {\cos x – \dfrac{1}{4}} \right)^2} \le \dfrac{{25}}{8}\\
\Leftrightarrow – \dfrac{{25}}{8} \le 2.{\left( {\cos x – \dfrac{1}{4}} \right)^2} – \dfrac{{25}}{8} \le 0\\
\Leftrightarrow – \dfrac{{25}}{8} \le y \le 0\\
\Leftrightarrow \left\{ \begin{array}{l}
GTNN:y = – \dfrac{{25}}{8}\,khi:\cos x = \dfrac{1}{4}\\
GTLN:y = 0\,khi:\cos x = – 1
\end{array} \right.
\end{array}$