let-f-x-ln-2-x-2-x-1-find-D-f-and-find-the-assymptotes-to-C-f-2-calculate-f-x-and-give-the-variation-of-f-3-give-the-graph-of-f-4-give-the-equation-of-tangent-to-C-f-at-po

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Question Number 40097 by maxmathsup by imad last updated on 15/Jul/18

$$letf\left(x\right)=ln\sqrt{\frac{2+x}{2-x}}$$$$1)find{D}_{f}andfindtheassymptotesto{C}_f$$$$2)calculate{f}^{{}^{\prime }}\left(x\right)andgivethevariationoff$$$$3)givethegraphoff$$$$4)givetheequationoftangentto{C}_{f}atpointE(\frac{1}{2},f\left(\frac{1}{2}\right))$$$$5)calculate{\int }_0^{1}f\left(x\right)dx.$$

Commented by math khazana by abdo last updated on 03/Aug/18

$$1)D_f=]-2,2[$$$${lim}_{x\to -2^{+}}f\left(x\right)=-\mathrm{\infty }$$$${lim}_{x\to 2^{-}}f\left(x\right)=+\mathrm{\infty }sox=-2andx=2are$$$$assymptotesto{C}_f$$$$2)wehavef\left(x\right)=\frac{1}{2}\{ln(2+x)-ln(2-x)\}\Rightarrow$$$$f^{{}^{\prime }}\left(x\right)=\frac{1}{2(2+x)}+\frac{1}{2(2-x)}=\frac{1}{2}\{\frac{1}{2+x}+\frac{1}{2-x}\}$$$$=\frac{1}{2}\frac{4}{4-x^2}=\frac{2}{4-x^2}>0because-2<x<2sof$$$$isincreasingon]-2,2[$$$$4)equ.oftangentisy=f^{{}^{\prime }}\left(\frac{1}{2}\right)(x-\frac{1}{2})+f\left(\frac{1}{2}\right)$$$$f^{{}^{\prime }}\left(\frac{1}{2}\right)=\frac{2}{4-\frac{1}{4}}=\frac{2\times 4}{15}=\frac{8}{15}$$$$f\left(\frac{1}{2}\right)=\frac{1}{2}ln\left(\frac{2+\frac{1}{2}}{2-\frac{1}{2}}\right)=\frac{1}{2}ln\left(\frac{5}{3}\right)\Rightarrow$$$$y=\frac{8}{15}(x-\frac{1}{2})+\frac{1}{2}ln\left(\frac{5}{3}\right)$$$$y=\frac{8}{15}x-\frac{4}{15}+\frac{1}{2}ln\left(\frac{5}{3}\right).$$

Commented by math khazana by abdo last updated on 03/Aug/18

$$5)letI={\int }_0^{1}ln\left(\sqrt{\frac{2+x}{2-x}}\right)dx$$$$I=\frac{1}{2}{\int }_0^{1}ln(2+x)dx-\frac{1}{2}{\int }_0^{1}ln(2-x)dxbut$$$${\int }_0^{1}ln(2+x)dx{=}_{2+x=t}{\int }_2^{3}ln\left(t\right)dt={[tln\left(t\right)-t]}_2^3$$$$=3ln\left(3\right)-3-2ln\left(2\right)+2=3ln\left(3\right)-2ln\left(2\right)-1$$$${\int }_0^{1}ln(2-x)dx{=}_{2-x=t}-{\int }_2^{1}ln\left(t\right)dt$$$$={\int }_1^{2}ln\left(t\right)dt={[tln\left(t\right)-t]}_1^2=2ln\left(2\right)-2+1$$$$=2ln\left(2\right)-1\Rightarrow$$$${\int }_0^{1}f\left(x\right)dx=\frac{3}{2}ln\left(3\right)-ln\left(2\right)-\frac{1}{2}-ln\left(2\right)+\frac{1}{2}$$$$=\frac{3}{2}ln\left(3\right)-2ln\left(2\right).$$

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