Menu Close

what-is-minimum-value-of-y-sin-x-cos-4-x-

Tinku Tara 0 Comments
Pin




Question Number 78767 by jagoll last updated on 20/Jan/20
what is minimum value of y = sin x+cos^4 x
$$\mathrm{what}\:\mathrm{is}\:\mathrm{minimum} \\ $$$$\mathrm{value}\:\mathrm{of}\:\mathrm{y}\:=\:\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:^{\mathrm{4}} \mathrm{x} \\ $$
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
$${since}\:\mathrm{cos}^{\mathrm{4}} \:{x}\geqslant\mathrm{0}, \\ $$$${y}_{{min}} =−\mathrm{1}\:{when}\:\mathrm{sin}\:{x}=−\mathrm{1}\:{and}\:\mathrm{cos}\:{x}=\mathrm{0} \\ $$
Commented by jagoll last updated on 20/Jan/20
why not use difffential sir?
$$\mathrm{why}\:\mathrm{not}\:\mathrm{use}\:\mathrm{difffential}\:\mathrm{sir}? \\ $$
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
$$\mathrm{if}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{sir}?\:\mathrm{equal}\:\mathrm{to} \\ $$$$\mathrm{1}\:\mathrm{when}\:\mathrm{cos}\:\mathrm{x}\:=\:\mathrm{1}\:\mathrm{and}\:\mathrm{sin}\:\mathrm{x}=\mathrm{0} \\ $$
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.
$${for}\:{minimum}\:{it}\:{is}\:{clear}: \\ $$$${y}=\mathrm{sin}\:{x}+\mathrm{cos}^{\mathrm{4}} \:{x} \\ $$$${y}_{{min}} \:{is}\:{when}\:{sin}\:{x}\:{is}\:{minimum}\:{which} \\ $$$${is}\:−\mathrm{1}\:{and}\:{at}\:{the}\:{same}\:{time}\:\mathrm{cos}^{\mathrm{4}} \:{x}\:{is}\: \\ $$$${minimum}\:{which}\:{is}\:\mathrm{0}.\:{it}\:{is}\:{possible} \\ $$$${that}\:{at}\:{the}\:{same}\:{time}\:\mathrm{sin}\:{x}=−\mathrm{1}\:{and} \\ $$$$\mathrm{cos}\:{x}=\mathrm{0}. \\ $$$${for}\:{maximum}\:{it}\:{is}\:{not}\:{so}\:{clear}: \\ $$$${y}=\mathrm{sin}\:{x}+\mathrm{cos}^{\mathrm{4}} \:{x} \\ $$$${since}\:{both}\:\mathrm{sin}\:{x}\:{and}\:\mathrm{cos}^{\mathrm{4}} \:{x}\:{can}\:{be} \\ $$$${positive},\:{but}\:{they}\:{can}\:{not}\:{be}\:\mathrm{1}\:{at}\:{the} \\ $$$${same}\:{time},\:{therefore}\:{y}_{{max}} \neq\mathrm{2}. \\ $$$$\mathrm{cos}\:{x}=\mathrm{1}\:{and}\:\mathrm{sin}\:{x}=\mathrm{0}\:{don}'{t}\:{give}\:{y}=\mathrm{1},\:{but} \\ $$$${not}\:{y}_{{max}} ,\:{for}\:{example}\:{with}\:{x}=\frac{\pi}{\mathrm{6}} \\ $$$${we}\:{have}\:\mathrm{sin}\:{x}=\frac{\mathrm{1}}{\mathrm{2}}\:{and}\:\mathrm{cos}\:{x}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{which} \\ $$$${give}\:{y}=\frac{\mathrm{1}}{\mathrm{2}}+\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{4}} =\mathrm{1}.\mathrm{0625}>\mathrm{1}.\: \\ $$$${to}\:{find}\:{where}\:{it}\:{is}\:{the}\:{y}_{{max}} \:{we}\:{need} \\ $$$${to}\:{do}\:{more}\:{investigation}\:{through} \\ $$$${y}'=\mathrm{0}. \\ $$
Commented by jagoll last updated on 20/Jan/20
yes sir. thanks you
$$\mathrm{yes}\:\mathrm{sir}.\:\mathrm{thanks}\:\mathrm{you} \\ $$
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
$${how}\:{to}\:{find}\:{maximum}: \\ $$$${y}=\mathrm{sin}\:{x}+\mathrm{cos}^{\mathrm{4}} \:{x} \\ $$$${y}'=\mathrm{cos}\:{x}−\mathrm{4}\:\mathrm{cos}^{\mathrm{3}} \:{x}\:\mathrm{sin}\:{x}=\mathrm{0} \\ $$$$\mathrm{cos}\:{x}\left(\mathrm{1}−\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:{x}\:\mathrm{sin}\:{x}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{cos}\:{x}=\mathrm{0}\:\Rightarrow….{clear}!\:\Rightarrow\mathrm{sin}\:{x}=\pm\mathrm{1}\:\Rightarrow{y}_{{min}} =−\mathrm{1},{y}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{1}−\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:{x}\:\mathrm{sin}\:{x}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{1}−\mathrm{4}\:\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:{x}\right)\mathrm{sin}\:{x}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{sin}^{\mathrm{3}} \:{x}−\mathrm{sin}\:{x}+\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{0} \\ $$$$\Delta=\left(−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{3}} +\left(\frac{\mathrm{1}}{\mathrm{8}}\right)^{\mathrm{2}} <\mathrm{0} \\ $$$$\Rightarrow\mathrm{sin}\:{x}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{8}}+\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\right)\:\:\left({k}=\mathrm{0},\mathrm{1},\mathrm{2}\right) \\ $$$${since}\:\mathrm{sin}\:{x}<\mathrm{1},\:{only}\:{k}=\mathrm{0}\:{and}\:\mathrm{1}\:{suitable}. \\ $$$${k}=\mathrm{0}:\:{y}=\mathrm{1}.\mathrm{129515} \\ $$$${k}=\mathrm{1}:\:{y}=\mathrm{0}.\mathrm{926658} \\ $$$${y}_{{max}} =\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{8}}\right)+\left\{\mathrm{1}−\left[\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{8}}\right)\right]^{\mathrm{2}} \right\}^{\mathrm{2}} \\ $$$$\approx\mathrm{1}.\mathrm{129515} \\ $$
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.
$$\mathrm{i}\:\mathrm{had}\:\mathrm{done}\:\mathrm{it}\:\mathrm{like}\:\mathrm{this}\:\mathrm{before}\::\: \\ $$$$\mathrm{let}\:\mathrm{t}=\:\mathrm{sin}\:\mathrm{x}\: \\ $$$$\mathrm{f}\left(\mathrm{t}\right)\:=\:\mathrm{t}\:+\:\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\frac{\mathrm{df}}{\mathrm{dt}}\:=\:\mathrm{1}\:−\mathrm{4t}\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\mathrm{4t}^{\mathrm{3}} −\mathrm{4t}+\mathrm{1}=\mathrm{0}\:\Rightarrow\:\mathrm{t}^{\mathrm{3}} −\mathrm{t}+\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{0} \\ $$$$\mathrm{i}\:\mathrm{have}\:\mathrm{trouble}\:\mathrm{with}\:\mathrm{the}\:\mathrm{equation} \\ $$$$.\:\mathrm{i}\:\mathrm{forget}\:\mathrm{the}\:\mathrm{cardano}\:\mathrm{formula}. \\ $$$$ \\ $$
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.
$${this}\:{leads}\:{to}\:{the}\:{same}\:{equation}\:{as}\:{mine}. \\ $$$${it}\:{has}\:{three}\:{roots}: \\ $$$${t}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{8}}+\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\right)\:\:\left({k}=\mathrm{0},\mathrm{1},\mathrm{2}\right) \\ $$$${but}\:{cardano}\:{won}'{t}\:{work}\:{in}\:{this}\:{case}. \\ $$
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) .
$${and}\:{your}\:{approach}\:{has}\:{a}\:{problem}: \\ $$$$\frac{{df}}{{dt}}=\mathrm{0}\:\Rightarrow\:{means}\:{only}\:{at}\:{t}=?\:{f}\:{is}\:{max}.\:{or} \\ $$$${min}. \\ $$$${but}\:{we}\:{want}\:{to}\:{know}\:{at}\:{x}=?\:{f}\:{is}\:{max}. \\ $$$${or}\:{min}. \\ $$$${correct}\:{is}\:{therefore}\:\frac{{df}}{{dx}}=\mathrm{0} \\ $$$$\Rightarrow\frac{{df}}{{dx}}=\frac{{df}}{{dt}}×\frac{{dt}}{{dx}}=\mathrm{0}\:\Rightarrow\frac{{df}}{{dt}}=\mathrm{0}\:{or}\:\frac{{dt}}{{dx}}=\mathrm{0} \\ $$$$\frac{{dt}}{{dx}}=\mathrm{0}\:{is}\:{vanished}\:{in}\:{your}\:{approach} \\ $$$${which}\:{means}\:\mathrm{cos}\:{x}=\mathrm{0},\:\Rightarrow\:{y}=\mathrm{1}\:{or} \\ $$$${y}=−\mathrm{1}={y}_{{min}} . \\ $$
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
$$\mathrm{i}\:\mathrm{don}'\mathrm{t}\:\mathrm{understand}\:\mathrm{how}\:\mathrm{to}\:\mathrm{get}\: \\ $$$$\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{8}}\right)+\frac{\mathrm{2k}\pi}{\mathrm{3}}\right)\:\mathrm{sir} \\ $$
Commented by jagoll last updated on 20/Jan/20
ow (df/dx)=(df/dt)×(dt/dx) it mean the function we consider two variables?
$$\mathrm{ow}\:\frac{\mathrm{df}}{\mathrm{dx}}=\frac{\mathrm{df}}{\mathrm{dt}}×\frac{\mathrm{dt}}{\mathrm{dx}}\:\mathrm{it}\:\mathrm{mean}\:\mathrm{the}\:\mathrm{function} \\ $$$$\mathrm{we}\:\mathrm{consider}\:\mathrm{two}\:\mathrm{variables}? \\ $$
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
$${f}={t}+\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} \:{and}\:{t}=\mathrm{sin}\:{x} \\ $$$$\frac{{df}}{{dx}}=\frac{{df}}{{dt}}×\frac{{dt}}{{dx}}=\left[\mathrm{1}−\mathrm{4}{t}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)\right]×\mathrm{cos}\:{x}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{cos}\:{x}=\mathrm{0}\:{or} \\ $$$$\Rightarrow{t}^{\mathrm{3}} −{t}+\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{0} \\ $$

Pin

Leave a Reply

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