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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))
$${what}\:{is}\:{minimum}\:{value} \\ $$$${of}\:{function}\:{f}\left({x}\right)= \\ $$$$\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}\:+\sqrt{{x}^{\mathrm{2}} −\mathrm{24}{x}+\mathrm{153}} \\ $$
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))))
$$\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}\geqslant\mathrm{2},\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{is}\:\mathrm{at}\:{x}=\mathrm{0}\:\Rightarrow \\ $$$$\Rightarrow\:\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}\:\mathrm{is}\:\mathrm{strictly}\:\mathrm{increasing}\:\mathrm{for}\:{x}\neq\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\mathrm{24}{x}+\mathrm{1}=\mathrm{0}\:\Leftrightarrow\:{x}=\mathrm{12}\pm\sqrt{\mathrm{143}}\:\Rightarrow \\ $$$$\Rightarrow\:\sqrt{{x}^{\mathrm{2}} −\mathrm{24}{x}+\mathrm{1}}\:\mathrm{is}\:\mathrm{not}\:\mathrm{defined}\:\mathrm{for} \\ $$$$\mathrm{12}−\sqrt{\mathrm{143}}<{x}<\mathrm{12}+\sqrt{\mathrm{143}},\:\mathrm{especially}\:\mathrm{not}\:\mathrm{definef} \\ $$$$\mathrm{for}\:{x}=\mathrm{0}\:\Rightarrow\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{of}\:\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}+\sqrt{{x}^{\mathrm{2}} −\mathrm{24}{x}+\mathrm{1}} \\ $$$$\mathrm{is}\:\mathrm{at}\:\mathrm{one}\:\mathrm{or}\:\mathrm{both}\:\mathrm{of}\:\mathrm{these}\:\mathrm{borders} \\ $$$${x}=\mathrm{12}−\sqrt{\mathrm{143}}\:\vee\:{x}=\mathrm{12}+\sqrt{\mathrm{143}}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{minimum}\:\mathrm{is}\:\mathrm{at}\:{x}=\mathrm{12}+\sqrt{\mathrm{143}}\:\mathrm{and}\:\mathrm{has}\:\mathrm{the} \\ $$$$\mathrm{value}\:\sqrt{\mathrm{291}−\mathrm{24}\sqrt{\mathrm{143}}} \\ $$
Commented by jagoll last updated on 10/Jan/20
sir why not use critical point from f ′(x)=0?
$${sir}\:{why}\:{not}\:{use}\:{critical}\:{point}\:{from} \\ $$$${f}\:'\left({x}\right)=\mathrm{0}? \\ $$
Commented by jagoll last updated on 10/Jan/20
my typo sir original problem f(x)=(√(x^2 +4)) + (√(x^2 −24x+153))
$${my}\:{typo}\:{sir}\:{original}\:{problem}\: \\ $$$${f}\left({x}\right)=\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}\:+\:\sqrt{{x}^{\mathrm{2}} −\mathrm{24}{x}+\mathrm{153}} \\ $$
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
$$\mathrm{because}\:{f}'\left({x}\right)=\mathrm{0}\:\mathrm{has}\:\mathrm{got}\:\mathrm{no}\:\mathrm{real}\:\mathrm{solution} \\ $$$${f}'\left({x}\right)=\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}}+\frac{{x}−\mathrm{12}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{24}{x}+\mathrm{1}}}=\mathrm{0} \\ $$$${x}\sqrt{{x}^{\mathrm{2}} −\mathrm{24}{x}+\mathrm{1}}=\left(\mathrm{12}−{x}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{4}} \\ $$$$\mathrm{squaring}\:\left[\mathrm{might}\:\mathrm{leads}\:\mathrm{to}\:\mathrm{false}\:\mathrm{solutions},\:\mathrm{we}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{must}\:\mathrm{check}\:\mathrm{them}!\right] \\ $$$${x}^{\mathrm{2}} \left({x}^{\mathrm{2}} −\mathrm{24}{x}+\mathrm{1}\right)=\left(\mathrm{12}−{x}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{4}\right) \\ $$$$\Rightarrow \\ $$$${x}^{\mathrm{2}} −\frac{\mathrm{32}}{\mathrm{49}}{x}+\frac{\mathrm{192}}{\mathrm{49}}=\mathrm{0} \\ $$$$\Rightarrow\:{x}=\frac{\mathrm{16}}{\mathrm{49}}\pm\frac{\mathrm{8}\sqrt{\mathrm{143}}}{\mathrm{49}}\mathrm{i} \\ $$$$\mathrm{both}\:\mathrm{solve}\:{f}'\left({x}\right)=\mathrm{0}\:\mathrm{but}\:\mathrm{both}\:\mathrm{are}\:\mathrm{not}\:\mathrm{real} \\ $$
Commented by MJS last updated on 10/Jan/20
anyway I hope you learned something new...
$$\mathrm{anyway}\:\mathrm{I}\:\mathrm{hope}\:\mathrm{you}\:\mathrm{learned}\:\mathrm{something}\:\mathrm{new}… \\ $$
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
$$\mathrm{with}\:\mathrm{the}\:\mathrm{new}\:\mathrm{equation}\:\mathrm{we}\:\mathrm{get} \\ $$$${f}'\left({x}\right)=\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}}+\frac{{x}−\mathrm{12}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{24}{x}+\mathrm{153}}}=\mathrm{0} \\ $$$${x}\sqrt{{x}^{\mathrm{2}} −\mathrm{24}{x}+\mathrm{153}}=\left(\mathrm{12}−{x}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{4}} \\ $$$$\mathrm{squaring}\:\mathrm{and}\:\mathrm{transforming} \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{96}}{\mathrm{5}}{x}−\frac{\mathrm{576}}{\mathrm{5}}=\mathrm{0} \\ $$$$\Rightarrow\:{x}=−\mathrm{24}\vee{x}=\frac{\mathrm{24}}{\mathrm{5}} \\ $$$${x}=−\mathrm{24}\:\mathrm{is}\:\mathrm{false};\:\mathrm{it}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{solve}\:{f}'\left({x}\right)=\mathrm{0} \\ $$$$\Rightarrow\:{x}=\frac{\mathrm{24}}{\mathrm{5}} \\ $$$$\Rightarrow\:\mathrm{minmum}\:\mathrm{at}\:{x}=\frac{\mathrm{24}}{\mathrm{5}};\:{f}\left(\frac{\mathrm{24}}{\mathrm{5}}\right)=\mathrm{13} \\ $$
Commented by jagoll last updated on 10/Jan/20
oo yes thanks you sir
$${oo}\:{yes}\:{thanks}\:{you}\:{sir} \\ $$

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