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a: \(f\left(x\right)=2x^2-7x+9\)
=>\(f'\left(x\right)=2\cdot2x-7=4x-7\)
Đặt f'(x)=0
=>\(4x-7=0\)
=>\(x=\dfrac{7}{4}\)
\(f\left(\dfrac{7}{4}\right)=2\cdot\left(\dfrac{7}{4}\right)^2-7\cdot\dfrac{7}{4}+9=\dfrac{23}{8}\)
\(f\left(-1\right)=2\left(-1\right)^2-7\cdot\left(-1\right)+9=18\)
\(f\left(4\right)=2\cdot4^2-7\cdot4+9=13\)
Vì \(f\left(\dfrac{7}{4}\right)< f\left(4\right)< f\left(-1\right)\)
nên \(f\left(x\right)_{max\left[-1;4\right]}=18;f\left(x\right)_{min\left[-1;4\right]}=\dfrac{23}{8}\)
b: \(f\left(x\right)=x^2+5x+3\)
=>\(f'\left(x\right)=2x+5\)
f'(x)=0
=>2x+5=0
=>2x=-5
=>\(x=-\dfrac{5}{2}\)
\(f\left(-\dfrac{5}{2}\right)=\left(-\dfrac{5}{2}\right)^2+5\cdot\dfrac{-5}{2}+3=\dfrac{25}{4}-\dfrac{25}{2}+3=-\dfrac{13}{4}\)
\(f\left(2\right)=2^2+5\cdot2+3=4+10+3=17\)
\(f\left(6\right)=6^2+5\cdot6+3=69\)
Vậy: \(f\left(x\right)_{max\left[2;6\right]}=69;f\left(x\right)_{min\left[2;6\right]}=-\dfrac{13}{4}\)
Áp dụng BĐT Cauchy cho cặp số dương \(\dfrac{1}{\left(z+x\right)};\dfrac{1}{\left(z+y\right)}\)
\(\dfrac{1}{\left(z+x\right)}+\dfrac{1}{\left(z+y\right)}\ge\dfrac{1}{2}.\dfrac{1}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\)
\(\Rightarrow\dfrac{xy}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\le\dfrac{2xy}{z+x}+\dfrac{2xy}{z+y}\left(1\right)\)
Tương tự ta được
\(\dfrac{zx}{\sqrt[]{\left(y+z\right)\left(y+x\right)}}\le\dfrac{2zx}{y+z}+\dfrac{2zx}{y+x}\left(2\right)\)
\(\dfrac{yz}{\sqrt[]{\left(x+y\right)\left(x+z\right)}}\le\dfrac{2yz}{x+y}+\dfrac{2yz}{x+z}\left(3\right)\)
\(\left(1\right)+\left(2\right)+\left(3\right)\) ta được :
\(P=\dfrac{yz}{\sqrt[]{\left(x+y\right)\left(x+z\right)}}+\dfrac{zx}{\sqrt[]{\left(y+z\right)\left(y+x\right)}}+\dfrac{xy}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\le\dfrac{2yz}{x+y}+\dfrac{2yz}{x+z}+\dfrac{2zx}{y+z}+\dfrac{2zx}{y+x}+\dfrac{2xy}{z+x}+\dfrac{2xy}{z+y}\)
\(\Rightarrow P\le2\left(x+y+z\right)=2.3=6\)
\(\Rightarrow GTLN\left(P\right)=6\left(tạix=y=z=1\right)\)
\(f\left(x\right)=\left(5+x\right)\left(x-6\right)\)
\(f\left(x\right)=30+x-x^2\)
\(f\left(x\right)=\frac{121}{4}-\left(x-\frac{1}{2}\right)^2\le\frac{121}{4}\left[\left(x-\frac{1}{2}\right)^2\ge0\right]\)
\(GTNN:f\left(x\right)=\frac{121}{4}\)
a/ \(0\le\sqrt{5-x^2}\le\sqrt{5}\)
Đặt \(t=\sqrt{5-x^2}\Rightarrow0\le t\le\sqrt{5}\)
\(y=-t^2-t+5\)
Ta có \(-\frac{b}{2a}=-\frac{1}{2}\notin\left[0;\sqrt{5}\right]\)
\(y\left(0\right)=5\) ; \(y\left(\sqrt{5}\right)=-\sqrt{5}\)
\(\Rightarrow y_{max}=5\) khi \(x=\pm\sqrt{5}\)
\(y_{min}=-\sqrt{5}\) khi \(x=0\)
Câu 2:
Nếu không thêm điều kiện gì thì cả min lẫn max đều ko tồn tại
Câu 3: Đề ko rõ
Câu 4: \(x>1\)
\(y=\frac{x-1}{20}+\frac{1}{2\sqrt{x-1}}+\frac{1}{2\sqrt{x-1}}+\frac{1}{20}\)
\(y\ge3\sqrt[3]{\frac{x-1}{80\left(x-1\right)}}+\frac{1}{20}=\frac{3}{2\sqrt[3]{10}}+\frac{1}{20}\)
Dấu "=" xảy ra khi \(\frac{x-1}{10}=\frac{1}{\sqrt{x-1}}\Rightarrow x=\sqrt[3]{100}+1\)