Question

[Hoàng]

Tìm GTNN của các hàm số sau trên miền đã chỉ ra.a) y=b)y=c)y=d)y=e)y=

Answer 1

[Hoàng Tử Hà]

Em chỉ thử thôi, giáo viên nào đi qua check hộ em với ạ!

\(y'=3-\dfrac{4}{x^3}\)

\(y'=0\Leftrightarrow x=\sqrt[3]{\dfrac{4}{3}}\notin[2;+\infty)\)

\(\Rightarrow y_{min}=f\left(2\right)=3.6+\dfrac{2}{2^2}=\dfrac{13}{2}\)

b/ \(y'=2-\dfrac{2}{x^3}\)

\(y'=0\Leftrightarrow x=1\notin(0;\dfrac{2}{3}]\)

\(f\left(0,4\right)=7,05;f\left(0,5\right)=5\Rightarrow ham-nghich-bien-trong-nua-khoang-(0;\dfrac{2}{3}]\)

\(\Rightarrow y_{min}=f\left(\dfrac{2}{3}\right)=2.\dfrac{2}{3}+\dfrac{1}{\left(\dfrac{2}{3}\right)^2}=\dfrac{43}{12}\)

c/ \(y=x+\dfrac{1}{x-1}\Rightarrow y'=1-\dfrac{1}{\left(x-1\right)^2}\)

\(y'=0\Leftrightarrow1-\dfrac{1}{\left(x-1\right)^2}=0\Leftrightarrow\left[{}\begin{matrix}x=0\notin\left(1;+\infty\right)\\x=2\in\left(1;+\infty\right)\end{matrix}\right.\)

\(f\left(\dfrac{3}{2}\right)=\dfrac{3}{2}+\dfrac{1}{\dfrac{3}{2}-1}=\dfrac{7}{2};f\left(2\right)=3;f\left(3\right)=\dfrac{7}{2}\)

=> ham nghich bien tren \(\left(1;2\right)\) va dong bien tren \([2;+\infty)\)

\(\Rightarrow y_{min}=f\left(2\right)=3\)

d/ \(y=\dfrac{1}{x}+\dfrac{2}{1-x}\Rightarrow y'=-\dfrac{1}{x^2}+\dfrac{2}{\left(1-x\right)^2}\)

\(y'=0\Leftrightarrow-\dfrac{1}{x^2}+\dfrac{2}{\left(1-x\right)^2}=0\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}-1\in\left(0;1\right)\\x=-1-\sqrt{2}\notin\left(0;1\right)\end{matrix}\right.\)

\(f\left(0,2\right)=\dfrac{15}{2};f\left(\sqrt{2}-1\right)=3+2\sqrt{2};f\left(0,5\right)=6\)

=> f(x) nghich bien tren \(\left(0;\sqrt{2}-1\right)\)

dong bien tren \([\sqrt{2}-1;1)\)

\(\Rightarrow y_{min}=f\left(\sqrt{2}-1\right)=3+2\sqrt{2}\)

e/ \(y=\dfrac{x^2+2x+2}{x+1}\Rightarrow y'=\dfrac{\left(x^2+2x+2\right)'\left(x+1\right)-\left(x+1\right)'\left(x^2+2x+2\right)}{\left(x+1\right)^2}\)

\(y'=\dfrac{\left(x+1\right).\left(2x+2\right)-x^2-2x-2}{\left(x+1\right)^2}=\dfrac{2x^2+4x+2-x^2-2x-2}{\left(x+1\right)^2}=\dfrac{x^2+2x}{x^2+2x+1}\)

\(y'=0\Leftrightarrow x^2+2x=0\Leftrightarrow x=0\in\left(-1;+\infty\right)\)

\(f\left(-0,5\right)=\dfrac{5}{2};f\left(0\right)=2;f\left(1\right)=\dfrac{5}{2}\)

=> f(x) nghich bien tren \(\left(-1;0\right)\)

dong bien tren \([0;+\infty)\)

\(\Rightarrow y_{min}=f\left(0\right)=2\)