| Q 179570 (Posted by Infinityaction 30.10.2022)
find the minimum of f(x)
f(x)=(√(x^2 +(4/x^2 )−8x−((12)/x)+25)) +(√(x^2 +(4/x^2 )−16x−((16)/x)+80))
−−−−−−−−−−−−−−−−−−
f(x)=(√((x−4)^2 +((2/x)−3)^2 )) +(√((x−8)^2 +((2/x)−4)^2 ))
Df=R−{0} f(x)≥0
Min(f(x))=x / f(x)=0
(√((x−4)^2 +((2/x)−3)^2 )) +(√((x−8)^2 +((2/x)−4)^2 )) =0
(x−4)^2 +((2/x)−3)^2 =(x−8)^2 +((2/x)−4)^2 (1)
x−8=(x−4)−4 and ((2/x)−4)=((2/x)−3)−1
x^2 −((55)/8)x+(1/2)=0 x=0,07351334 and x=6,80148666}
x=0,073513334
f(x)=49,044679(rejete)
x=6,80148666 f(x)=7,7899055
Donc 7,7899055 est minimum de f(x)
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