Q 179570 (Posted by Infinityaction 30.10.2022)
find the minimum of f(x)
f(x)=(β(x^2 +(4/x^2 )β8xβ((12)/x)+25)) +(β(x^2 +(4/x^2 )β16xβ((16)/x)+80))
ββββββββββββββββββ
f(x)=(β((xβ4)^2 +((2/x)β3)^2 )) +(β((xβ8)^2 +((2/x)β4)^2 ))
Df=Rβ{0} f(x)β₯0
Min(f(x))=x / f(x)=0
(β((xβ4)^2 +((2/x)β3)^2 )) +(β((xβ8)^2 +((2/x)β4)^2 )) =0
(xβ4)^2 +((2/x)β3)^2 =(xβ8)^2 +((2/x)β4)^2 (1)
xβ8=(xβ4)β4 and ((2/x)β4)=((2/x)β3)β1
x^2 β((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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