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Question Number 61495 by bhanukumarb2@gmail.com last updated on 03/Jun/19

Commented by bhanukumarb2@gmail.com last updated on 03/Jun/19

$$provesecondinwhichbookicanget$$$$thesetypeapproximation$$

Commented by bhanukumarb2@gmail.com last updated on 03/Jun/19

$$provebymvt$$

Commented by tanmay last updated on 03/Jun/19

$$wehavetoprove(\frac{x^2}{2}+cosx)>1$$$$fromgraphitisclear...$$$$$$

Commented by tanmay last updated on 03/Jun/19

Commented by tanmay last updated on 03/Jun/19

Commented by tanmay last updated on 03/Jun/19

Commented by tanmay last updated on 03/Jun/19

Answered by MJS last updated on 03/Jun/19

$$f\left(x\right)=\cos x\mathrm{has}\mathrm{a}\mathrm{local}\mathrm{maximum}\mathrm{at}x=0$$$$g\left(x\right)=1-\frac{x^2}{2}\mathrm{has}\mathrm{an}\mathrm{absolute}\mathrm{maximum}\mathrm{ar}x=0$$$$f\left(0\right)=g\left(0\right)=1$$$$$$$$\mathrm{the}\mathrm{curvature}\mathrm{of}\mathrm{a}\mathrm{given}\mathrm{function}u\mathrm{is}$$$$\frac{u^{″}}{{(1+u^{\prime 2})}^{\frac{3}{2}}}$$$$\mathrm{the}\mathrm{curvature}\mathrm{of}f\left(x\right)\mathrm{is}$$$$c_f\left(x\right)=-\frac{\cos x}{{(1+{\sin }^{2}x)}^{\frac{3}{2}}}$$$$\mathrm{the}\mathrm{curvature}\mathrm{of}g\left(x\right)\mathrm{is}$$$$c_g\left(x\right)=-\frac{1}{{(1+x^2)}^{\frac{3}{2}}}$$$$c_f\left(0\right)=c_g\left(0\right)=-1$$$$c_f^{\prime }\left(0\right)=c_g^{\prime }\left(0\right)=0$$$$c_f^{″}\left(0\right)=4$$$$c_g^{″}\left(0\right)=3$$$$\Rightarrow c_f^{\prime }\mathrm{is}\mathrm{changing}\mathrm{faster}\mathrm{than}{c}_g^{\prime }$$$$\Rightarrow c_f\mathrm{is}\mathrm{bent}\mathrm{harder}\mathrm{towards}y=0\mathrm{than}{c}_g$$$$\Rightarrow f\mathrm{is}\mathrm{bent}\mathrm{less}\mathrm{than}g$$$$\Rightarrow \mathrm{because}\mathrm{we}\mathrm{know}\mathrm{both}\mathrm{share}\mathrm{the}$$$$\mathrm{maximum}\mathrm{at}\left(\begin{array}{c}0\\ 1\end{array}\right)\Rightarrow f⩾g$$

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