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Question Number 122293 by mathocean1 last updated on 15/Nov/20

$$givenf\left(x\right)={cos}^{2}x$$$$forx\in [-\frac{\pi }{12};\frac{\pi }{12}],\mid f^{\prime }\left(x\right)\mid ⩽\frac{1}{2}.$$$$1)showthat\mathrm{\forall }x,y\in [-\frac{\pi }{12};\frac{\pi }{12}];$$$$\mid {cos}^{2}x-{cos}^{2}y\mid ⩽\frac{1}{2}\mid x-y\mid$$

Commented by ZiYangLee last updated on 16/Nov/20

$$\mathrm{By}\mathrm{MVT},\mathrm{we}\mathrm{have}$$$$\frac{f\left(x\right)-f\left(y\right)}{x-y}=f^{\prime }\left(c\right)⩽\frac{1}{2}$$$$\mathrm{where}c\in (x,y),$$$$\mathrm{Hence},$$$$\mid f\left(x\right)-f\left(y\right)\mid ⩽\frac{1}{2}\mid x-y\mid$$

Answered by mathmax by abdo last updated on 15/Nov/20

$$\mathrm{\forall }\mathrm{x}\in [-\frac{\pi }{12},\frac{\pi }{12}]\mid {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)\mid ⩽\frac{1}{2}\Rightarrow -\frac{1}{2}⩽{\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)⩽\frac{1}{2}\Rightarrow$$$$-\frac{1}{2}{\int }_{\mathrm{x}}^{\mathrm{y}}\mathrm{dt}⩽{\int }_{\mathrm{x}}^{\mathrm{y}}{\mathrm{f}}^{{}^{\prime }}\left(\mathrm{t}\right)\mathrm{dt}⩽\frac{1}{2}{\int }_{\mathrm{x}}^{\mathrm{y}}\mathrm{dt}\Rightarrow$$$$-\frac{1}{2}(\mathrm{y}-\mathrm{x})⩽\mathrm{f}\left(\mathrm{y}\right)-\mathrm{f}\left(\mathrm{x}\right)⩽\frac{1}{2}(\mathrm{y}-\mathrm{x})\Rightarrow \mid \mathrm{f}\left(\mathrm{x}\right)-\mathrm{f}\left(\mathrm{y}\right)\mid ⩽\frac{1}{2}\mid \mathrm{x}-\mathrm{y}\mid \Rightarrow$$$$\mid {\cos }^2\mathrm{x}-{\cos }^2\mathrm{y}\mid ⩽\frac{\mid \mathrm{x}-\mathrm{y}\mid }{2}$$

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