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Question-161802




Question Number 161802 by femKaren last updated on 22/Dec/21
Answered by mr W last updated on 22/Dec/21
f(x)=cos^3  x+sin^3  x  f′(x)=−3 cos^2  x sin x+3 sin^2  x cos x            =3 cos x sin x (sin x−cos x)=0  for x∈ (0,(π/2)): cos x≠0, sin x≠0  therefore  sin x−cos x=0   ⇒tan x=1 ⇒x=(π/4) ⇒cos x=sin x=((√2)/2)  f(x)_(extremum) =2×(((√2)/2))^3 =((√2)/2)
$${f}\left({x}\right)=\mathrm{cos}^{\mathrm{3}} \:{x}+\mathrm{sin}^{\mathrm{3}} \:{x} \\ $$$${f}'\left({x}\right)=−\mathrm{3}\:\mathrm{cos}^{\mathrm{2}} \:{x}\:\mathrm{sin}\:{x}+\mathrm{3}\:\mathrm{sin}^{\mathrm{2}} \:{x}\:\mathrm{cos}\:{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{3}\:\mathrm{cos}\:{x}\:\mathrm{sin}\:{x}\:\left(\mathrm{sin}\:{x}−\mathrm{cos}\:{x}\right)=\mathrm{0} \\ $$$${for}\:{x}\in\:\left(\mathrm{0},\frac{\pi}{\mathrm{2}}\right):\:\mathrm{cos}\:{x}\neq\mathrm{0},\:\mathrm{sin}\:{x}\neq\mathrm{0} \\ $$$${therefore} \\ $$$$\mathrm{sin}\:{x}−\mathrm{cos}\:{x}=\mathrm{0}\: \\ $$$$\Rightarrow\mathrm{tan}\:{x}=\mathrm{1}\:\Rightarrow{x}=\frac{\pi}{\mathrm{4}}\:\Rightarrow\mathrm{cos}\:{x}=\mathrm{sin}\:{x}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$${f}\left({x}\right)_{{extremum}} =\mathrm{2}×\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{3}} =\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$
Commented by femKaren last updated on 22/Dec/21
thank you so much!
$${thank}\:{you}\:{so}\:{much}! \\ $$

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