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Question Number 178624 by infinityaction last updated on 19/Oct/22

$$\mathbf{let}\mathbf{f}:[0,1]\to \mathbb{R}\mathbf{be}\mathbf{given}\mathbf{by}$$$$\mathbf{f}\left(\mathbf{x}\right)=\frac{{(1+{\mathbf{x}}^{\frac{1}{3}})}^3+{(1-{\mathbf{x}}^{\frac{1}{3}})}^3}{8(1+\mathbf{x})}\mathbf{then}$$$$\mathbf{max}\{\mathbf{f}\left(\mathbf{x}\right):\mathbf{x}\in [0,1]\}-\mathbf{min}\{\mathbf{f}\left(\mathbf{x}\right):\mathbf{x}\in [0,1]\}$$$$\mathrm{is}$$

Answered by a.lgnaoui last updated on 20/Oct/22

$$\mathrm{posons}={\mathrm{x}}^{\frac{1}{3}}=\mathrm{z};\mathrm{x}={\mathrm{z}}^3$$$${(1+\mathrm{z})}^3+{(1-\mathrm{z})}^3=2[{(1+\mathrm{z})}^2+{(1-\mathrm{z})}^2-(1-{\mathrm{z}}^2)]=2(1+{3\mathrm{z}}^2)$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{1+{3\mathrm{z}}^2}{4(1+{\mathrm{z}}^3)}=\frac{1+{3\mathrm{x}}^{\frac{2}{3}}}{4(1+\mathrm{x})}$$$$\frac{\mathrm{df}}{\mathrm{dx}}=\frac{1}{4}\left[\frac{3\frac{2}{3}{\mathrm{x}}^{\frac{-1}{3}}(1+\mathrm{x})-(1+{3\mathrm{x}}^{\frac{2}{3}})}{{(1+\mathrm{x})}^2}\right]=\frac{{2\mathrm{x}}^{\frac{-1}{3}}-{\mathrm{x}}^{\frac{2}{3}}-1}{4{(1+\mathrm{x})}^2}$$$$(\frac{2}{\mathrm{z}}-{\mathrm{z}}^2-1)\times \frac{1}{4{(1+z^3)}^2}=\frac{2-{\mathrm{z}}^3-\mathrm{z}}{4\mathrm{z}(1+{\mathrm{z}}^3)}$$$$-\frac{{\mathrm{z}}^3+\mathrm{z}-2}{4\mathrm{z}(1+{\mathrm{z}}^3)}=-\frac{{(\mathrm{z}-1)}^2(\mathrm{z}+2)}{4\mathrm{z}(1+{\mathrm{z}}^3)}\Rightarrow [\mathrm{ex}t\mathrm{remum}(x=1f\left(1\right)=\frac{1}{2}=max\left(f\left(x\right)\right);f\left(0\right)=\frac{1}{4}=min\left(f\left(x\right)\right)]$$$$max\left(f\left(x\right)\right)-min\left(f\left(x\right)\right)=\frac{1}{4}$$$$$$

Commented by infinityaction last updated on 20/Oct/22

$$thankssir$$

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