Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

Question Number 79644 by loveineq. last updated on 26/Jan/20

f(x) = ((4(4x^2 +3))/(4x^2 +4x+5)) prove that f(x) ≥ 2 .

$${f}\left({x}\right)\:=\:\frac{\mathrm{4}\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{3}\right)}{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{5}} \\ $$$$\mathrm{prove}\:\mathrm{that}\:{f}\left({x}\right)\:\geqslant\:\mathrm{2}\:. \\ $$

Commented by john santu last updated on 27/Jan/20

let f(x)<2 ((16x^2 +12)/(4x^2 +4x+5))<2 ⇒4x^2 +4x+5>0 ∀x∈R 16x^2 +12<8x^2 +8x+10 8x^2 −8x+2<0 ⇒Δ= 64−4.8.2=0 x^2 −x+(1/4)=(x−(1/2))^2 ≥0 contradiction with the initial statement. so f(x)≥2.

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)<\mathrm{2} \\ $$$$\frac{\mathrm{16x}^{\mathrm{2}} +\mathrm{12}}{\mathrm{4x}^{\mathrm{2}} +\mathrm{4x}+\mathrm{5}}<\mathrm{2}\:\Rightarrow\mathrm{4x}^{\mathrm{2}} +\mathrm{4x}+\mathrm{5}>\mathrm{0}\:\forall\mathrm{x}\in\mathbb{R} \\ $$$$\mathrm{16x}^{\mathrm{2}} +\mathrm{12}<\mathrm{8x}^{\mathrm{2}} +\mathrm{8x}+\mathrm{10} \\ $$$$\mathrm{8x}^{\mathrm{2}} −\mathrm{8x}+\mathrm{2}<\mathrm{0}\:\Rightarrow\Delta=\:\mathrm{64}−\mathrm{4}.\mathrm{8}.\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\frac{\mathrm{1}}{\mathrm{4}}=\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$\mathrm{contradiction}\:\mathrm{with}\:\mathrm{the}\:\mathrm{initial} \\ $$$$\mathrm{statement}. \\ $$$$\mathrm{so}\:\mathrm{f}\left(\mathrm{x}\right)\geqslant\mathrm{2}. \\ $$$$ \\ $$

Answered by mr W last updated on 27/Jan/20

f(x) = ((4(4x^2 +3))/(4x^2 +4x+5)) = ((4(4x^2 +4x+5−4x−2))/(4x^2 +4x+5)) =4(1−((x+(1/2))/(x^2 +x+(5/4)))) =4(1−((x+(1/2))/((x+(1/2))^2 +1))) =4(1−(1/((x+(1/2))+(1/((x+(1/2))))))) =4(1+(1/(t+(1/t))))≥4(1−(1/2))=2 with t=−x−(1/2)

$${f}\left({x}\right)\:=\:\frac{\mathrm{4}\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{3}\right)}{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{5}} \\ $$$$=\:\frac{\mathrm{4}\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{5}−\mathrm{4}{x}−\mathrm{2}\right)}{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{5}} \\ $$$$=\mathrm{4}\left(\mathrm{1}−\frac{{x}+\frac{\mathrm{1}}{\mathrm{2}}}{{x}^{\mathrm{2}} +{x}+\frac{\mathrm{5}}{\mathrm{4}}}\right) \\ $$$$=\mathrm{4}\left(\mathrm{1}−\frac{{x}+\frac{\mathrm{1}}{\mathrm{2}}}{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{1}}\right) \\ $$$$=\mathrm{4}\left(\mathrm{1}−\frac{\mathrm{1}}{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)}}\right) \\ $$$$=\mathrm{4}\left(\mathrm{1}+\frac{\mathrm{1}}{{t}+\frac{\mathrm{1}}{{t}}}\right)\geqslant\mathrm{4}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{2} \\ $$$${with}\:{t}=−{x}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com