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f-x-13-12-x-25-24-1-x-find-Min-f-




Question Number 210587 by mnjuly1970 last updated on 13/Aug/24
       f(x)= (√( 13 −12(√x)  )) + (√(25 −24(√(1−x)) ))          find :    Min ( f )=?
$$ \\ $$$$\:\:\:\:\:{f}\left({x}\right)=\:\sqrt{\:\mathrm{13}\:−\mathrm{12}\sqrt{{x}}\:\:}\:+\:\sqrt{\mathrm{25}\:−\mathrm{24}\sqrt{\mathrm{1}−{x}}\:} \\ $$$$ \\ $$$$\:\:\:\:\:\:{find}\::\:\:\:\:\mathrm{M}{in}\:\left(\:{f}\:\right)=? \\ $$$$ \\ $$$$ \\ $$
Commented by Frix last updated on 13/Aug/24
I get 2(√5)
$$\mathrm{I}\:\mathrm{get}\:\mathrm{2}\sqrt{\mathrm{5}} \\ $$
Commented by mnjuly1970 last updated on 13/Aug/24
yes sir....2(√5) is correct.
$${yes}\:{sir}….\mathrm{2}\sqrt{\mathrm{5}}\:{is}\:{correct}. \\ $$
Commented by Ghisom last updated on 14/Aug/24
f(x)=(√(2p^2 +2p+1−2p(p+1)(√x)))+(√(2p^2 +6p+5−2(p+1)(p+2)(√(1−x))))  p≥0  ⇒  min (f(x)) =(√(2(p^2 +2p+2)))  at x=((p^2 (p^3 +5p^2 +12p+10−(p+2)^2 (√(p^4 +4p^3 +10p^2 +12p+4))))/(2(p+1)^2 (p^2 +2p+2)^2 ))
$${f}\left({x}\right)=\sqrt{\mathrm{2}{p}^{\mathrm{2}} +\mathrm{2}{p}+\mathrm{1}−\mathrm{2}{p}\left({p}+\mathrm{1}\right)\sqrt{{x}}}+\sqrt{\mathrm{2}{p}^{\mathrm{2}} +\mathrm{6}{p}+\mathrm{5}−\mathrm{2}\left({p}+\mathrm{1}\right)\left({p}+\mathrm{2}\right)\sqrt{\mathrm{1}−{x}}} \\ $$$${p}\geqslant\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\mathrm{min}\:\left({f}\left({x}\right)\right)\:=\sqrt{\mathrm{2}\left({p}^{\mathrm{2}} +\mathrm{2}{p}+\mathrm{2}\right)} \\ $$$$\mathrm{at}\:{x}=\frac{{p}^{\mathrm{2}} \left({p}^{\mathrm{3}} +\mathrm{5}{p}^{\mathrm{2}} +\mathrm{12}{p}+\mathrm{10}−\left({p}+\mathrm{2}\right)^{\mathrm{2}} \sqrt{{p}^{\mathrm{4}} +\mathrm{4}{p}^{\mathrm{3}} +\mathrm{10}{p}^{\mathrm{2}} +\mathrm{12}{p}+\mathrm{4}}\right)}{\mathrm{2}\left({p}+\mathrm{1}\right)^{\mathrm{2}} \left({p}^{\mathrm{2}} +\mathrm{2}{p}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$

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