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2-x-1-5x-x-2-4-x-24-




Question Number 140273 by liberty last updated on 06/May/21
2(√(x−1)) +5x = (√((x^2 +4)(x+24)))
$$\mathrm{2}\sqrt{\mathrm{x}−\mathrm{1}}\:+\mathrm{5x}\:=\:\sqrt{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{4}\right)\left(\mathrm{x}+\mathrm{24}\right)} \\ $$
Answered by MJS_new last updated on 06/May/21
x=5  I saw that (5^2 +4)(5+24)=29^2     f(x)=5x+2(√(x−1))−(√((x^2 +4)(x+24)))  (df/dx)=0  5+(1/( (√(x−1))))−((3x^2 +48x+4)/( 2(√((x^2 +4)(x+24)))))=0  ⇒ x=5 ⇒ maximum at x=5 ⇒ no other solution
$${x}=\mathrm{5} \\ $$$$\mathrm{I}\:\mathrm{saw}\:\mathrm{that}\:\left(\mathrm{5}^{\mathrm{2}} +\mathrm{4}\right)\left(\mathrm{5}+\mathrm{24}\right)=\mathrm{29}^{\mathrm{2}} \\ $$$$ \\ $$$${f}\left({x}\right)=\mathrm{5}{x}+\mathrm{2}\sqrt{{x}−\mathrm{1}}−\sqrt{\left({x}^{\mathrm{2}} +\mathrm{4}\right)\left({x}+\mathrm{24}\right)} \\ $$$$\frac{{df}}{{dx}}=\mathrm{0} \\ $$$$\mathrm{5}+\frac{\mathrm{1}}{\:\sqrt{{x}−\mathrm{1}}}−\frac{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{48}{x}+\mathrm{4}}{\:\mathrm{2}\sqrt{\left({x}^{\mathrm{2}} +\mathrm{4}\right)\left({x}+\mathrm{24}\right)}}=\mathrm{0} \\ $$$$\Rightarrow\:{x}=\mathrm{5}\:\Rightarrow\:\mathrm{maximum}\:\mathrm{at}\:{x}=\mathrm{5}\:\Rightarrow\:\mathrm{no}\:\mathrm{other}\:\mathrm{solution} \\ $$
Answered by liberty last updated on 06/May/21

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