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Question Number 138085 by liberty last updated on 10/Apr/21

Given a curve y = (1/(x^2 +1)). Find the equation of tangent line with have slope of tangent minimum .

$${Given}\:{a}\:{curve}\:{y}\:=\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}. \\ $$$${Find}\:{the}\:{equation}\:{of}\:{tangent} \\ $$$${line}\:{with}\:{have}\:{slope}\:{of}\:{tangent} \\ $$$${minimum}\:. \\ $$

Answered by EDWIN88 last updated on 10/Apr/21

let the eq of tangent line is y=px+q where p = minimum value of y′ step(1) y = (x^2 +1)^(−1) ; y′=−2x(x^2 +1)^(−2) we want p(x)=((−2x)/((x^2 +1)^2 )) minimum p′(x)= ((−2(x^2 +1)^2 −(−2x)(4x)(x^2 +1))/((x^2 +1)^4 )) p′(x)= ((−2(x^2 +1)+8x^2 )/((x^2 +1)^3 )) = 0 ; p′(x)=((6x^2 −2)/((x^2 +1)^3 )) ⇒ 6x^2 = 2 ; x^2 =(1/3) or x= ± ((√3)/3) chek p′′(x)= ((12x(x^2 +1)^3 −6x(6x^2 −2)(x^2 +1)^2 )/((x^2 +1)^6 )) p′′(x)=((24x(1−x^2 ))/((x^2 +1)^4 )) >0 when x=(1/( (√3))) so minimum of p(x) when x= (1/( (√3))) then p=((−2((1/( (√3)))))/((16)/9)) = −((2(√3))/3)×(9/(16)) = −((3(√3))/8) (2) contact point of tangent ((1/( (√3))) , (3/4)) thus eq of tangent line have the minimum slope of curve y=(1/(x^2 +1)) is 3(√3) x +8y = 3(√3)((1/( (√3))))+8((3/4)) 3(√3) x +8y = 9

$$\:{let}\:{the}\:{eq}\:{of}\:{tangent}\:{line}\:{is}\:{y}={px}+{q} \\ $$$${where}\:{p}\:=\:{minimum}\:{value}\:{of}\:{y}' \\ $$$${step}\left(\mathrm{1}\right)\:{y}\:=\:\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{−\mathrm{1}} \:;\:{y}'=−\mathrm{2}{x}\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{−\mathrm{2}} \\ $$$${we}\:{want}\:{p}\left({x}\right)=\frac{−\mathrm{2}{x}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\:{minimum} \\ $$$${p}'\left({x}\right)=\:\frac{−\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −\left(−\mathrm{2}{x}\right)\left(\mathrm{4}{x}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{4}} } \\ $$$${p}'\left({x}\right)=\:\frac{−\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{1}\right)+\mathrm{8}{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }\:=\:\mathrm{0}\:;\:{p}'\left({x}\right)=\frac{\mathrm{6}{x}^{\mathrm{2}} −\mathrm{2}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$\Rightarrow\:\mathrm{6}{x}^{\mathrm{2}} \:=\:\mathrm{2}\:;\:{x}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{3}}\:{or}\:{x}=\:\pm\:\frac{\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$${chek}\:{p}''\left({x}\right)=\:\frac{\mathrm{12}{x}\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} −\mathrm{6}{x}\left(\mathrm{6}{x}^{\mathrm{2}} −\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{6}} } \\ $$$${p}''\left({x}\right)=\frac{\mathrm{24}{x}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{4}} }\:>\mathrm{0}\:{when}\:{x}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$${so}\:{minimum}\:{of}\:{p}\left({x}\right)\:{when}\:{x}=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$${then}\:{p}=\frac{−\mathrm{2}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)}{\frac{\mathrm{16}}{\mathrm{9}}}\:=\:−\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}×\frac{\mathrm{9}}{\mathrm{16}}\:=\:−\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{8}} \\ $$$$\left(\mathrm{2}\right)\:{contact}\:{point}\:{of}\:{tangent}\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:,\:\frac{\mathrm{3}}{\mathrm{4}}\right) \\ $$$${thus}\:{eq}\:{of}\:{tangent}\:{line}\:{have}\:{the}\:{minimum} \\ $$$${slope}\:{of}\:{curve}\:{y}=\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\:{is}\: \\ $$$$\:\:\:\mathrm{3}\sqrt{\mathrm{3}}\:{x}\:+\mathrm{8}{y}\:=\:\mathrm{3}\sqrt{\mathrm{3}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)+\mathrm{8}\left(\frac{\mathrm{3}}{\mathrm{4}}\right) \\ $$$$\:\:\mathrm{3}\sqrt{\mathrm{3}}\:{x}\:+\mathrm{8}{y}\:=\:\mathrm{9}\: \\ $$

Commented by EDWIN88 last updated on 10/Apr/21

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