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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+x2−24x+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+4⩾2,theminimumisatx=0⇒⇒x2+4isstrictlyincreasingforx≠0x2−24x+1=0⇔x=12±143⇒⇒x2−24x+1isnotdefinedfor12−143<x<12+143,especiallynotdefinefforx=0⇒theminimumofx2+4+x2−24x+1isatoneorbothofthesebordersx=12−143∨x=12+143⇒⇒minimumisatx=12+143andhasthevalue291−24143
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+x2−24x+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+x−12x2−24x+1=0xx2−24x+1=(12−x)x2+4squaring[mightleadstofalsesolutions,wemustcheckthem!]x2(x2−24x+1)=(12−x)2(x2+4)⇒x2−3249x+19249=0⇒x=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+x−12x2−24x+153=0xx2−24x+153=(12−x)x2+4squaringandtransformingx2+965x−5765=0⇒x=−24∨x=245x=−24isfalse;itdoesn′tsolvef′(x)=0⇒x=245⇒minmumatx=245;f(245)=13
Commented by jagoll last updated on 10/Jan/20
oo yes thanks you sir
ooyesthanksyousir

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