Menu Close

Given-a-curve-y-1-x-2-1-Find-the-equation-of-tangent-line-with-have-slope-of-tangent-minimum-




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 .
Givenacurvey=1x2+1.Findtheequationoftangentlinewithhaveslopeoftangentminimum.
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
lettheeqoftangentlineisy=px+qwherep=minimumvalueofystep(1)y=(x2+1)1;y=2x(x2+1)2wewantp(x)=2x(x2+1)2minimump(x)=2(x2+1)2(2x)(4x)(x2+1)(x2+1)4p(x)=2(x2+1)+8x2(x2+1)3=0;p(x)=6x22(x2+1)36x2=2;x2=13orx=±33chekp(x)=12x(x2+1)36x(6x22)(x2+1)2(x2+1)6p(x)=24x(1x2)(x2+1)4>0whenx=13sominimumofp(x)whenx=13thenp=2(13)169=233×916=338(2)contactpointoftangent(13,34)thuseqoftangentlinehavetheminimumslopeofcurvey=1x2+1is33x+8y=33(13)+8(34)33x+8y=9
Commented by EDWIN88 last updated on 10/Apr/21

Leave a Reply

Your email address will not be published. Required fields are marked *