Question Number 196309 by sonukgindia last updated on 22/Aug/23
Commented by Frix last updated on 22/Aug/23
$$\sqrt{\mathrm{65}}\:\mathrm{at}\:{x}=−\frac{\mathrm{13}}{\mathrm{11}} \\ $$
Answered by AST last updated on 22/Aug/23
![For x<−2∨x>0;f(x) is increasing ⇒min{f(x)} occurs at −2<x<0 ((2x+1)/( (√(2x^2 +2x+13))))+((2x+4)/( (√(2x^2 +8x+26))))=0 ⇒(2x+1)((√(2x^2 +8x+26)))+(2x+4)((√(2x^2 +2x+13)))=0 (2x+1)^2 (2x^2 +8x+26)=(2x+4)^2 (2x^2 +2x+13) ⇒(2x+1)^2 [(2x+4)^2 +36]=(2x+4)^2 [(2x+1)^2 +25] ⇒[6(2x+1)]^2 =[5(2x+4)]^2 ⇒12x+6=10x+20 or 12x+6=−10x−20⇒x=3 or x=((−13)/(11)) ⇒min{f(x)}=f(((−13)/(11)))=(√(65))](https://www.tinkutara.com/question/Q196316.png)
$${For}\:{x}<−\mathrm{2}\vee{x}>\mathrm{0};{f}\left({x}\right)\:{is}\:{increasing} \\ $$$$\Rightarrow{min}\left\{{f}\left({x}\right)\right\}\:{occurs}\:{at}\:−\mathrm{2}<{x}<\mathrm{0} \\ $$$$\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{13}}}+\frac{\mathrm{2}{x}+\mathrm{4}}{\:\sqrt{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{8}{x}+\mathrm{26}}}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{2}{x}+\mathrm{1}\right)\left(\sqrt{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{8}{x}+\mathrm{26}}\right)+\left(\mathrm{2}{x}+\mathrm{4}\right)\left(\sqrt{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{13}}\right)=\mathrm{0} \\ $$$$\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{8}{x}+\mathrm{26}\right)=\left(\mathrm{2}{x}+\mathrm{4}\right)^{\mathrm{2}} \left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{13}\right) \\ $$$$\Rightarrow\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} \left[\left(\mathrm{2}{x}+\mathrm{4}\right)^{\mathrm{2}} +\mathrm{36}\right]=\left(\mathrm{2}{x}+\mathrm{4}\right)^{\mathrm{2}} \left[\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{25}\right] \\ $$$$\Rightarrow\left[\mathrm{6}\left(\mathrm{2}{x}+\mathrm{1}\right)\right]^{\mathrm{2}} =\left[\mathrm{5}\left(\mathrm{2}{x}+\mathrm{4}\right)\right]^{\mathrm{2}} \Rightarrow\mathrm{12}{x}+\mathrm{6}=\mathrm{10}{x}+\mathrm{20} \\ $$$${or}\:\mathrm{12}{x}+\mathrm{6}=−\mathrm{10}{x}−\mathrm{20}\Rightarrow{x}=\mathrm{3}\:{or}\:{x}=\frac{−\mathrm{13}}{\mathrm{11}} \\ $$$$\Rightarrow{min}\left\{{f}\left({x}\right)\right\}={f}\left(\frac{−\mathrm{13}}{\mathrm{11}}\right)=\sqrt{\mathrm{65}} \\ $$
Answered by mr W last updated on 22/Aug/23
![f(x)=(√(2(x+(1/2))^2 +((25)/2)))+(√(2(x+2)^2 +18)) f(x)=(√2)[(√((−x−(1/2))^2 +((5/2))^2 ))+(√((x+2)^2 +3^2 ))] ≥(√2)×(√((−x−(1/2)+x+2)^2 +((5/2)+3)^2 )) =(√((3^2 +11^2 )/2))=(√(65))=minimum when (−x−(1/2)):(x+2)=(5/2):3, i.e. x=−((13)/(11))](https://www.tinkutara.com/question/Q196318.png)
$${f}\left({x}\right)=\sqrt{\mathrm{2}\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{25}}{\mathrm{2}}}+\sqrt{\mathrm{2}\left({x}+\mathrm{2}\right)^{\mathrm{2}} +\mathrm{18}} \\ $$$${f}\left({x}\right)=\sqrt{\mathrm{2}}\left[\sqrt{\left(−{x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} }+\sqrt{\left({x}+\mathrm{2}\right)^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} }\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\geqslant\sqrt{\mathrm{2}}×\sqrt{\left(−{x}−\frac{\mathrm{1}}{\mathrm{2}}+{x}+\mathrm{2}\right)^{\mathrm{2}} +\left(\frac{\mathrm{5}}{\mathrm{2}}+\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:=\sqrt{\frac{\mathrm{3}^{\mathrm{2}} +\mathrm{11}^{\mathrm{2}} }{\mathrm{2}}}=\sqrt{\mathrm{65}}={minimum} \\ $$$${when}\:\left(−{x}−\frac{\mathrm{1}}{\mathrm{2}}\right):\left({x}+\mathrm{2}\right)=\frac{\mathrm{5}}{\mathrm{2}}:\mathrm{3},\:{i}.{e}. \\ $$$${x}=−\frac{\mathrm{13}}{\mathrm{11}} \\ $$