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Question Number 77810 by jagoll last updated on 10/Jan/20
what is minimum value  of function f(x)=  (√(x^2 +4)) +(√(x^2 −24x+153))
whatisminimumvalueoffunctionf(x)=x2+4+x224x+153
Answered by MJS last updated on 10/Jan/20
(√(x^2 +4))≥2, the minimum is at x=0 ⇒  ⇒ (√(x^2 +4)) is strictly increasing for x≠0  x^2 −24x+1=0 ⇔ x=12±(√(143)) ⇒  ⇒ (√(x^2 −24x+1)) is not defined for  12−(√(143))<x<12+(√(143)), especially not definef  for x=0 ⇒ the minimum of (√(x^2 +4))+(√(x^2 −24x+1))  is at one or both of these borders  x=12−(√(143)) ∨ x=12+(√(143)) ⇒  ⇒ minimum is at x=12+(√(143)) and has the  value (√(291−24(√(143))))
x2+42,theminimumisatx=0x2+4isstrictlyincreasingforx0x224x+1=0x=12±143x224x+1isnotdefinedfor12143<x<12+143,especiallynotdefinefforx=0theminimumofx2+4+x224x+1isatoneorbothofthesebordersx=12143x=12+143minimumisatx=12+143andhasthevalue29124143
Commented by jagoll last updated on 10/Jan/20
sir why not use critical point from  f ′(x)=0?
sirwhynotusecriticalpointfromf(x)=0?
Commented by jagoll last updated on 10/Jan/20
my typo sir original problem   f(x)=(√(x^2 +4)) + (√(x^2 −24x+153))
mytyposiroriginalproblemf(x)=x2+4+x224x+153
Commented by MJS last updated on 10/Jan/20
because f′(x)=0 has got no real solution  f′(x)=(x/( (√(x^2 +4))))+((x−12)/( (√(x^2 −24x+1))))=0  x(√(x^2 −24x+1))=(12−x)(√(x^2 +4))  squaring [might leads to false solutions, we                       must check them!]  x^2 (x^2 −24x+1)=(12−x)^2 (x^2 +4)  ⇒  x^2 −((32)/(49))x+((192)/(49))=0  ⇒ x=((16)/(49))±((8(√(143)))/(49))i  both solve f′(x)=0 but both are not real
becausef(x)=0hasgotnorealsolutionf(x)=xx2+4+x12x224x+1=0xx224x+1=(12x)x2+4squaring[mightleadstofalsesolutions,wemustcheckthem!]x2(x224x+1)=(12x)2(x2+4)x23249x+19249=0x=1649±814349ibothsolvef(x)=0butbotharenotreal
Commented by MJS last updated on 10/Jan/20
anyway I hope you learned something new...
anywayIhopeyoulearnedsomethingnew
Answered by MJS last updated on 10/Jan/20
with the new equation we get  f′(x)=(x/( (√(x^2 +4))))+((x−12)/( (√(x^2 −24x+153))))=0  x(√(x^2 −24x+153))=(12−x)(√(x^2 +4))  squaring and transforming  x^2 +((96)/5)x−((576)/5)=0  ⇒ x=−24∨x=((24)/5)  x=−24 is false; it doesn′t solve f′(x)=0  ⇒ x=((24)/5)  ⇒ minmum at x=((24)/5); f(((24)/5))=13
withthenewequationwegetf(x)=xx2+4+x12x224x+153=0xx224x+153=(12x)x2+4squaringandtransformingx2+965x5765=0x=24x=245x=24isfalse;itdoesntsolvef(x)=0x=245minmumatx=245;f(245)=13
Commented by jagoll last updated on 10/Jan/20
oo yes thanks you sir
ooyesthanksyousir

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