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Question Number 100771 by Rio Michael last updated on 28/Jun/20

$$\mathrm{sho}\mathrm{that}(0,\frac{1}{2})\mathrm{is}\mathrm{a}\mathrm{point}\mathrm{of}\mathrm{symetry}\mathrm{for}\mathrm{the}\mathrm{curve}$$$$f\left(x\right)=x+\frac{1}{1-e^x}$$$$\mathrm{Please}\mathrm{make}\mathrm{a}\mathrm{reference}\mathrm{to}\mathrm{a}\mathrm{book}\mathrm{i}\mathrm{can}\mathrm{understand}$$$$\mathrm{centre}\mathrm{of}\mathrm{symmetry}\mathrm{of}\mathrm{rational}\mathrm{functions}\mathrm{and}\mathrm{functions}$$$$\mathrm{like}\mathrm{this}$$

Commented by abdomathmax last updated on 28/Jun/20

$$\mathrm{all}\mathrm{vuibert}\mathrm{and}\mathrm{ellipces}\mathrm{books}\mathrm{are}\mathrm{good}...$$

Answered by MJS last updated on 28/Jun/20

$$\left(\begin{array}{c}p\\ q\end{array}\right)\mathrm{is}\mathrm{the}\mathrm{symmetry}\mathrm{center}\mathrm{of}f\left(x\right)$$$$\iff$$$$q-f(p-x)=-(q-f\left(x\right))$$$$$$$$\mathrm{in}\mathrm{our}\mathrm{case}$$$$\frac{1}{2}-(1-x-\frac{1}{1-{\mathrm{e}}^x})=-(\frac{1}{2}-x-\frac{1}{1-{\mathrm{e}}^x})$$$$\mathrm{is}\mathrm{true}$$

Commented by MJS last updated on 28/Jun/20

$$\mathrm{sorry}\mathrm{I}\mathrm{know}\mathrm{no}\mathrm{books}\mathrm{on}\mathrm{this}\mathrm{topic}...$$

Answered by abdomathmax last updated on 28/Jun/20

$$\mathrm{I}(\mathrm{a},\mathrm{b})\mathrm{is}\mathrm{centre}\mathrm{of}\mathrm{symetry}\iff \mathrm{f}(2\mathrm{a}-\mathrm{x})=2\mathrm{b}-\mathrm{f}\left(\mathrm{x}\right)$$$$\mathrm{I}(0,\frac{1}{2})\mathrm{we}\mathrm{must}\mathrm{prove}\mathrm{f}(2\times 0-\mathrm{x})=2\times \frac{1}{2}-\mathrm{f}\left(\mathrm{x}\right)\Rightarrow$$$$\mathrm{f}(-\mathrm{x})=1-\mathrm{f}\left(\mathrm{x}\right)\mathrm{we}\mathrm{have}$$$$\mathrm{f}(-\mathrm{x})=-\mathrm{x}+\frac{1}{1-{\mathrm{e}}^{-\mathrm{x}}}=-\mathrm{x}+\frac{{\mathrm{e}}^{\mathrm{x}}}{{\mathrm{e}}^{\mathrm{x}}-1}=\frac{(1-\mathrm{x}){\mathrm{e}}^{\mathrm{x}}+\mathrm{x}}{{\mathrm{e}}^{\mathrm{x}}-1}$$$$1-\mathrm{f}\left(\mathrm{x}\right)=1-\mathrm{x}-\frac{1}{1-{\mathrm{e}}^{\mathrm{x}}}=\frac{1-{\mathrm{e}}^{\mathrm{x}}-\mathrm{x}+{\mathrm{xe}}^{\mathrm{x}}-1}{1-{\mathrm{e}}^{\mathrm{x}}}$$$$=\frac{-\mathrm{x}+(x-1)e^x}{1-e^x}=\frac{x+(1-x)e^x}{e^x-1}$$$$tbeequalityisproved..$$

Commented by Rio Michael last updated on 28/Jun/20

$$\mathrm{thank}\mathrm{you}\mathrm{all}$$

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