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Question Number 78767 by jagoll last updated on 20/Jan/20
what is minimum value of y = sin x+cos^4 x
whatisminimumvalueofy=sinx+cos4x
Commented by mr W last updated on 20/Jan/20
since cos^4 x≥0, y_(min) =−1 when sin x=−1 and cos x=0
sincecos4x⩾0,ymin=−1whensinx=−1andcosx=0
Commented by jagoll last updated on 20/Jan/20
why not use difffential sir?
whynotusedifffentialsir?
Commented by jagoll last updated on 20/Jan/20
if the maximum value sir? equal to 1 when cos x = 1 and sin x=0
ifthemaximumvaluesir?equalto1whencosx=1andsinx=0
Commented by mr W last updated on 20/Jan/20
for minimum it is clear: y=sin x+cos^4 x y_(min) is when sin x is minimum which is −1 and at the same time cos^4 x is minimum which is 0. it is possible that at the same time sin x=−1 and cos x=0. for maximum it is not so clear: y=sin x+cos^4 x since both sin x and cos^4 x can be positive, but they can not be 1 at the same time, therefore y_(max) ≠2. cos x=1 and sin x=0 don′t give y=1, but not y_(max) , for example with x=(π/6) we have sin x=(1/2) and cos x=((√3)/2) which give y=(1/2)+(((√3)/2))^4 =1.0625>1. to find where it is the y_(max) we need to do more investigation through y′=0.
forminimumitisclear:y=sinx+cos4xyminiswhensinxisminimumwhichis−1andatthesametimecos4xisminimumwhichis0.itispossiblethatatthesametimesinx=−1andcosx=0.formaximumitisnotsoclear:y=sinx+cos4xsincebothsinxandcos4xcanbepositive,buttheycannotbe1atthesametime,thereforeymax≠2.cosx=1andsinx=0don′tgivey=1,butnotymax,forexamplewithx=π6wehavesinx=12andcosx=32whichgivey=12+(32)4=1.0625>1.tofindwhereitistheymaxweneedtodomoreinvestigationthroughy′=0.
Commented by jagoll last updated on 20/Jan/20
yes sir. thanks you
yessir.thanksyou
Commented by mr W last updated on 20/Jan/20
how to find maximum: y=sin x+cos^4 x y′=cos x−4 cos^3 x sin x=0 cos x(1−4 cos^2 x sin x)=0 ⇒cos x=0 ⇒....clear! ⇒sin x=±1 ⇒y_(min) =−1,y=1 ⇒1−4 cos^2 x sin x=0 ⇒1−4 (1−sin^2 x)sin x=0 ⇒sin^3 x−sin x+(1/4)=0 Δ=(−(1/3))^3 +((1/8))^2 <0 ⇒sin x=(2/( (√3))) sin ((1/3)sin^(−1) ((3(√3))/8)+((2kπ)/3)) (k=0,1,2) since sin x<1, only k=0 and 1 suitable. k=0: y=1.129515 k=1: y=0.926658 y_(max) =(2/( (√3))) sin ((1/3)sin^(−1) ((3(√3))/8))+{1−[(2/( (√3))) sin ((1/3)sin^(−1) ((3(√3))/8))]^2 }^2 ≈1.129515
howtofindmaximum:y=sinx+cos4xy′=cosx−4cos3xsinx=0cosx(1−4cos2xsinx)=0⇒cosx=0⇒….clear!⇒sinx=±1⇒ymin=−1,y=1⇒1−4cos2xsinx=0⇒1−4(1−sin2x)sinx=0⇒sin3x−sinx+14=0Δ=(−13)3+(18)2<0⇒sinx=23sin(13sin−1338+2kπ3)(k=0,1,2)sincesinx<1,onlyk=0and1suitable.k=0:y=1.129515k=1:y=0.926658ymax=23sin(13sin−1338)+{1−[23sin(13sin−1338)]2}2≈1.129515
Commented by mr W last updated on 20/Jan/20
Commented by jagoll last updated on 20/Jan/20
i had done it like this before : let t= sin x f(t) = t + (1−t^2 )^2 (df/dt) = 1 −4t(1−t^2 )=0 4t^3 −4t+1=0 ⇒ t^3 −t+(1/4)=0 i have trouble with the equation . i forget the cardano formula.
ihaddoneitlikethisbefore:lett=sinxf(t)=t+(1−t2)2dfdt=1−4t(1−t2)=04t3−4t+1=0⇒t3−t+14=0ihavetroublewiththeequation.iforgetthecardanoformula.
Commented by mr W last updated on 20/Jan/20
this leads to the same equation as mine. it has three roots: t=(2/( (√3))) sin ((1/3)sin^(−1) ((3(√3))/8)+((2kπ)/3)) (k=0,1,2) but cardano won′t work in this case.
thisleadstothesameequationasmine.ithasthreeroots:t=23sin(13sin−1338+2kπ3)(k=0,1,2)butcardanowon′tworkinthiscase.
Commented by mr W last updated on 20/Jan/20
and your approach has a problem: (df/dt)=0 ⇒ means only at t=? f is max. or min. but we want to know at x=? f is max. or min. correct is therefore (df/dx)=0 ⇒(df/dx)=(df/dt)×(dt/dx)=0 ⇒(df/dt)=0 or (dt/dx)=0 (dt/dx)=0 is vanished in your approach which means cos x=0, ⇒ y=1 or y=−1=y_(min) .
andyourapproachhasaproblem:dfdt=0⇒meansonlyatt=?fismax.ormin.butwewanttoknowatx=?fismax.ormin.correctisthereforedfdx=0⇒dfdx=dfdt×dtdx=0⇒dfdt=0ordtdx=0dtdx=0isvanishedinyourapproachwhichmeanscosx=0,⇒y=1ory=−1=ymin.
Commented by jagoll last updated on 20/Jan/20
i don′t understand how to get (2/( (√3))) sin ((1/3)sin^(−1) (((3(√3))/8))+((2kπ)/3)) sir
idon′tunderstandhowtoget23sin(13sin−1(338)+2kπ3)sir
Commented by jagoll last updated on 20/Jan/20
ow (df/dx)=(df/dt)×(dt/dx) it mean the function we consider two variables?
owdfdx=dfdt×dtdxitmeanthefunctionweconsidertwovariables?
Commented by mr W last updated on 20/Jan/20
f=t+(1−t^2 )^2 and t=sin x (df/dx)=(df/dt)×(dt/dx)=[1−4t(1−t^2 )]×cos x=0 ⇒cos x=0 or ⇒t^3 −t+(1/4)=0
f=t+(1−t2)2andt=sinxdfdx=dfdt×dtdx=[1−4t(1−t2)]×cosx=0⇒cosx=0or⇒t3−t+14=0

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