\(y=\sqrt{5-4x}\)
Txd : \(\left\{{}\begin{matrix}5-4x\ge0\\-1\le x\le1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{5}{4}\\-1\le x\le1\end{matrix}\right.\) \(\Leftrightarrow-1\le x\le\dfrac{5}{4}\)
\(\Rightarrow Txd:D=\left[-1;\dfrac{5}{4}\right]\)
\(y'=\dfrac{-4}{2\sqrt{5-4x}}=\dfrac{-2}{\sqrt{5-4x}}< 0,\forall x\in D\)
Nên hàm số cho nghịch biến trên \(\left[-1;\dfrac{5}{4}\right]\)
\(\Rightarrow y\left(max\right)=y\left(-1\right)=\sqrt{5-4.\left(-1\right)}=3\)
\(\)\(y\left(min\right)=y\left(\dfrac{5}{4}\right)=\sqrt{5-4.\dfrac{5}{4}}=0\)
\(y'=-\dfrac{4}{2\sqrt{5-4x}}< 0\) nên hàm ko có cực trị
\(y\left(-1\right)=\sqrt{5-4.\left(-1\right)}=3\) ; \(y\left(1\right)=\sqrt{5-4.1}=1\)
\(\Rightarrow\min\limits_{\left[-1;1\right]}y=y\left(1\right)=1\) ; \(\max\limits_{\left[-1;1\right]}y=y\left(-1\right)=3\)
