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Max-amp-min-value-of-function-f-x-6-x-12-x-




Question Number 153956 by liberty last updated on 12/Sep/21
 Max & min value of function   f(x)=(√(6−x)) +(√(12+x)) .
Max&minvalueoffunctionf(x)=6x+12+x.
Answered by EDWIN88 last updated on 12/Sep/21
 domain of f(x) = [−12,6 ]  by Chaucy −Schward inequality  2[(6−x)+(12+x) ]≥ ((√(6−x))+(√(12+x)) )^2   2×18 ≥ [ (√(6−x)) +(√(12+x)) ]^2   ⇒(√(18)) ≤ (√(6−x)) +(√(12+x)) ≤ (√(36))  ⇒3(√2) ≤ (√(6−x)) +(√(12+x)) ≤ 6    { ((min=3(√2))),((max=6)) :}
domainoff(x)=[12,6]byChaucySchwardinequality2[(6x)+(12+x)](6x+12+x)22×18[6x+12+x]2186x+12+x36326x+12+x6{min=32max=6
Commented by liberty last updated on 13/Sep/21
yes correct
yescorrect
Answered by ajfour last updated on 15/Sep/21
let  x−3=t  f(t)=(√(9−t))+(√(9+t))  ⇒     t∈[−9,9]  f^( 2) (t)=18+2(√(81−t^2 ))  clearly f_(min) =3(√2),  f_(max) =6
letx3=tf(t)=9t+9+tt[9,9]f2(t)=18+281t2clearlyfmin=32,fmax=6

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