Question

Tìm GTLN, GTNN của hàm số: \(y=\frac{sinx+cosx-1}{cosx+3}\)

Answer 1

Lời giải:

$y=\frac{\sin x+\cos x-1}{\cos x+3}$
$\Rightarrow y(\cos x+3)-\sin x-\cos x+1=0$

$\Leftrightarrow -\sin x+(y-1)\cos x=-3y-1$

Áp dụng BĐT Bunhiacopxky:

$(-3y-1)^2=[-\sin x+(y-1)\cos x]^2\leq (\sin ^2x+\cos ^2x)[1+(y-1)^2]$

$\Leftrightarrow (-3y-1)^2\leq 1+(y-1)^2$
$\Leftrightarrow 8y^2+8y-1\leq 0$

$\Leftrightarrow \frac{-2-\sqrt{6}}{4}\leq y\leq \frac{-2+\sqrt{6}}{4}$

Vậy $y_{\max}=\frac{-2+\sqrt{6}}{4}$

$y_{\min}=\frac{-2-\sqrt{6}}{4}$