Question Number 196343 by universe last updated on 23/Aug/23
$$\:\:\mathrm{find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{x}\right) \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:=\:\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{2x}\:+\mathrm{5}}\:\:+\:\:\sqrt{\mathrm{4x}^{\mathrm{2}} \:−\mathrm{4x}\:+\mathrm{10}\:} \\ $$
Answered by MM42 last updated on 23/Aug/23
$$\sqrt{\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{4}}+\sqrt{\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{9}} \\ $$$$\Rightarrow{min}\sim\mathrm{5} \\ $$$$ \\ $$
Answered by Frix last updated on 23/Aug/23
![f(x)=(√(x^2 −2x+5))+2(√(x^2 −x+(5/2))) Solving f′(x)=0 leads to: x≈.635144580 f(x)≈5.04515892 [No useful exact solution possible] Maybe check the question...](https://www.tinkutara.com/question/Q196346.png)
$${f}\left({x}\right)=\sqrt{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{5}}+\mathrm{2}\sqrt{{x}^{\mathrm{2}} −{x}+\frac{\mathrm{5}}{\mathrm{2}}} \\ $$$$\mathrm{Solving}\:{f}'\left({x}\right)=\mathrm{0}\:\mathrm{leads}\:\mathrm{to}: \\ $$$${x}\approx.\mathrm{635144580} \\ $$$${f}\left({x}\right)\approx\mathrm{5}.\mathrm{04515892} \\ $$$$\left[\mathrm{No}\:\mathrm{useful}\:\mathrm{exact}\:\mathrm{solution}\:\mathrm{possible}\right] \\ $$$$\mathrm{Maybe}\:\mathrm{check}\:\mathrm{the}\:\mathrm{question}… \\ $$