Given-the-function-f-x-x-3-x-2-and-g-x-1-2-xe-x-1-Find-the-centre-of-symmetry-of-f-2-Define-the-monotony-of-g-and-if-possible-draw-a-variation-table-for-g-x-3-Sketch-the-funct

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Question Number 107454 by Rio Michael last updated on 10/Aug/20

$$\mathrm{Given}\mathrm{the}\mathrm{function}f\left(x\right)=\frac{x+3}{x-2}\mathrm{and}\mathrm{g}\left(x\right)=\frac{1}{2}{xe}^x$$$$\left(1\right)\mathrm{Find}\mathrm{the}\mathrm{centre}\mathrm{of}\mathrm{symmetry}\mathrm{of}f.$$$$\left(2\right)\mathrm{Define}\mathrm{the}\mathrm{monotony}\mathrm{of}\mathrm{g}\mathrm{and}\mathrm{if}\mathrm{possible}\mathrm{draw}\mathrm{a}\mathrm{variation}$$$$\mathrm{table}\mathrm{for}\mathrm{g}\left(x\right).$$$$\left(3\right)\mathrm{Sketch}\mathrm{the}\mathrm{function}\mathrm{g}\left(x\right)$$$$\left(4\right)\mathrm{determine}\mathrm{if}f\mathrm{and}\mathrm{g}\mathrm{intersect}.$$

Answered by abdomsup last updated on 11/Aug/20

$$f\left(x\right)=\frac{x+3}{x-2}=\frac{x-2+5}{x-2}=1+\frac{5}{x-2}$$$$\Rightarrow f\left(x\right)-1=\frac{5}{x-2}letw(2,1)$$$$letprovethatf(4-x)=2-f\left(x\right)$$$$(f(2a-x)=2b-f\left(x\right))$$$$f(4-x)=\frac{4-x+3}{4-x-2}=\frac{7-x}{2-x}=\frac{x-7}{x-2}$$$$2-f\left(x\right)=2-\frac{x+3}{x-2}=\frac{2x-4-x-3}{x-2}$$$$=\frac{x-7}{x-2}\Rightarrow f(4-x)=2-f\left(x\right)\Rightarrow$$$$w(2,1)isacentreofsymetry$$$$$$

Commented by Rio Michael last updated on 11/Aug/20

$$\mathrm{Thank}\mathrm{you}\mathrm{so}\mathrm{much}\mathrm{sir}.$$

Answered by abdomsup last updated on 11/Aug/20

$$2)g\left(x\right)=\frac{1}{2}x{e}^x$$$${lim}_{x\to -\mathrm{\infty }}g\left(x\right)=0$$$${lim}_{x\to +\mathrm{\infty }}g\left(x\right)=+\mathrm{\infty }$$$$g^{{}^{\prime }}\left(x\right)=\frac{1}{2}{e}^x+\frac{1}{2}{xe}^x=\frac{1}{2}(x+1)e^x$$$$g^{{}^{\prime }}\left(x\right)⩾0\Rightarrow x⩾-1vsruation$$$$x-\mathrm{\infty }-1+\mathrm{\infty }$$$$g^{{}^{\prime }}-+$$$$g0decr-\frac{1}{2e}incr+\mathrm{\infty }$$$$$$

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