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Question Number 196343 by universe last updated on 23/Aug/23

$$\mathrm{find}\mathrm{the}\mathrm{minimum}\mathrm{value}\mathrm{of}\mathrm{f}\left(\mathrm{x}\right)$$$$\mathrm{f}\left(\mathrm{x}\right)=\sqrt{{\mathrm{x}}^2-2\mathrm{x}+5}+\sqrt{{4\mathrm{x}}^2-4\mathrm{x}+10}$$

Answered by MM42 last updated on 23/Aug/23

$$\sqrt{{(x-1)}^2+4}+\sqrt{{(2x-1)}^2+9}$$$$\Rightarrow min\sim 5$$$$$$

Answered by Frix last updated on 23/Aug/23

$$f\left(x\right)=\sqrt{x^2-2x+5}+2\sqrt{x^2-x+\frac{5}{2}}$$$$\mathrm{Solving}{f}^{\prime }\left(x\right)=0\mathrm{leads}\mathrm{to}:$$$$x\approx .635144580$$$$f\left(x\right)\approx 5.04515892$$$$\left[\mathrm{No}\mathrm{useful}\mathrm{exact}\mathrm{solution}\mathrm{possible}\right]$$$$\mathrm{Maybe}\mathrm{check}\mathrm{the}\mathrm{question}...$$

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