Question Number 178624 by infinityaction last updated on 19/Oct/22
![let f:[0,1]→ R be given by f(x) = (((1+x^(1/3) )^3 +(1−x^(1/3) )^3 )/(8(1+x))) then max{f(x): x∈[0,1]}−min{f(x):x∈[0,1]} is](https://www.tinkutara.com/question/Q178624.png)
$$\:\:\:\:\:\:\:\boldsymbol{\mathrm{let}}\:\boldsymbol{\mathrm{f}}:\left[\mathrm{0},\mathrm{1}\right]\rightarrow\:\mathbb{R}\:\boldsymbol{\mathrm{be}}\:\boldsymbol{\mathrm{given}}\:\boldsymbol{\mathrm{by}} \\ $$$$\:\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)\:=\:\:\frac{\left(\mathrm{1}+\boldsymbol{\mathrm{x}}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)^{\mathrm{3}} +\left(\mathrm{1}−\boldsymbol{\mathrm{x}}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)^{\mathrm{3}} }{\mathrm{8}\left(\mathrm{1}+\boldsymbol{\mathrm{x}}\right)}\:\:\:\boldsymbol{\mathrm{then}} \\ $$$$\:\:\boldsymbol{\mathrm{max}}\left\{\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right):\:\boldsymbol{\mathrm{x}}\in\left[\mathrm{0},\mathrm{1}\right]\right\}−\boldsymbol{\mathrm{min}}\left\{\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right):\boldsymbol{\mathrm{x}}\in\left[\mathrm{0},\mathrm{1}\right]\right\} \\ $$$$\mathrm{is} \\ $$
Answered by a.lgnaoui last updated on 20/Oct/22
![posons=x^(1/3) =z ; x=z^3 (1+z)^3 +(1−z)^3 =2[(1+z)^2 +(1−z)^2 −(1−z^2 )]=2(1+3z^2 ) f(x)=((1+3z^2 )/(4(1+z^3 )))=((1+3x^(2/3) )/(4(1+x))) (df/dx)=(1/4)[((3(2/3)x^((−1)/3) (1+x)−(1+3x^(2/3) ))/((1+x)^2 ))]=((2x^((−1)/3) −x^(2/3) −1)/(4(1+x)^2 )) ((2/z) −z^2 −1)×(1/(4(1+z^3 )^2 ))=((2−z^3 −z)/(4z(1+z^3 ))) −((z^3 +z−2)/(4z(1+z^3 ))) =−(((z−1)^2 (z+2))/(4z(1+z^3 ))) ⇒[extremum(x=1 f(1)=(1/2)=max(f(x)) ; f(0)=(1/4)=min(f(x))] max(f(x))−min(f(x))=(1/4)](https://www.tinkutara.com/question/Q178705.png)
$$\mathrm{posons}=\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{3}}} =\mathrm{z}\:\:\:\:\:\:;\:\mathrm{x}=\mathrm{z}^{\mathrm{3}} \\ $$$$\left(\mathrm{1}+\mathrm{z}\right)^{\mathrm{3}} +\left(\mathrm{1}−\mathrm{z}\right)^{\mathrm{3}} =\mathrm{2}\left[\left(\mathrm{1}+\mathrm{z}\right)^{\mathrm{2}} +\left(\mathrm{1}−\mathrm{z}\right)^{\mathrm{2}} −\left(\mathrm{1}−\mathrm{z}^{\mathrm{2}} \right)\right]=\mathrm{2}\left(\mathrm{1}+\mathrm{3z}^{\mathrm{2}} \right) \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{1}+\mathrm{3z}^{\mathrm{2}} }{\mathrm{4}\left(\mathrm{1}+\mathrm{z}^{\mathrm{3}} \right)}=\frac{\mathrm{1}+\mathrm{3x}^{\frac{\mathrm{2}}{\mathrm{3}}} }{\mathrm{4}\left(\mathrm{1}+\mathrm{x}\right)} \\ $$$$\frac{\mathrm{df}}{\mathrm{dx}}=\frac{\mathrm{1}}{\mathrm{4}}\left[\frac{\mathrm{3}\frac{\mathrm{2}}{\mathrm{3}}\mathrm{x}^{\frac{−\mathrm{1}}{\mathrm{3}}} \left(\mathrm{1}+\mathrm{x}\right)−\left(\mathrm{1}+\mathrm{3x}^{\frac{\mathrm{2}}{\mathrm{3}}} \right)}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2}} }\right]=\frac{\mathrm{2x}^{\frac{−\mathrm{1}}{\mathrm{3}}} −\mathrm{x}^{\frac{\mathrm{2}}{\mathrm{3}}} −\mathrm{1}}{\mathrm{4}\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2}} } \\ $$$$\left(\frac{\mathrm{2}}{\mathrm{z}}\:−\mathrm{z}^{\mathrm{2}} −\mathrm{1}\right)×\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{1}+{z}^{\mathrm{3}} \right)^{\mathrm{2}} }=\frac{\mathrm{2}−\mathrm{z}^{\mathrm{3}} −\mathrm{z}}{\mathrm{4z}\left(\mathrm{1}+\mathrm{z}^{\mathrm{3}} \right)} \\ $$$$−\frac{\mathrm{z}^{\mathrm{3}} +\mathrm{z}−\mathrm{2}}{\mathrm{4z}\left(\mathrm{1}+\mathrm{z}^{\mathrm{3}} \right)}\:=−\frac{\left(\mathrm{z}−\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{z}+\mathrm{2}\right)}{\mathrm{4z}\left(\mathrm{1}+\mathrm{z}^{\mathrm{3}} \right)}\:\:\:\:\Rightarrow\left[\mathrm{ex}{t}\mathrm{remum}\left({x}=\mathrm{1}\:\:{f}\left(\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{2}}={max}\left({f}\left({x}\right)\right)\:;\:\:{f}\left(\mathrm{0}\right)=\frac{\mathrm{1}}{\mathrm{4}}={min}\left({f}\left({x}\right)\right)\right]\right. \\ $$$${max}\left({f}\left({x}\right)\right)−{min}\left({f}\left({x}\right)\right)=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$ \\ $$
Commented by infinityaction last updated on 20/Oct/22
$${thanks}\:{sir} \\ $$