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Question Number 116376 by bemath last updated on 03/Oct/20

$$\mathrm{Let}\mathrm{f}\mathrm{be}\mathrm{a}\mathrm{function}\mathrm{defined}\mathrm{on}\mathrm{non}\mathrm{zero}$$$$\mathrm{real}\mathrm{numbers}\mathrm{such}\mathrm{that}\frac{27\mathrm{f}(-\mathrm{x})}{\mathrm{x}}-{\mathrm{x}}^2\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)=-{2\mathrm{x}}^2$$$$\mathrm{for}\mathrm{\forall }\mathrm{x}\ne 0.\mathrm{Find}\to \{\begin{array}{l}\mathrm{f}\left(\mathrm{x}\right)\\ \mathrm{f}\left(3\right)\end{array}?$$

Answered by bobhans last updated on 03/Oct/20

$$\mathrm{Letting}\mathrm{x}=-\mathrm{y},\mathrm{we}\mathrm{get}$$$$-\frac{27\mathrm{f}\left(\mathrm{y}\right)}{\mathrm{y}}-{\mathrm{y}}^2\mathrm{f}(-\frac{1}{\mathrm{y}})=-{2\mathrm{y}}^2...\left(1\right)$$$$\mathrm{Letting}\mathrm{x}=\frac{1}{\mathrm{y}},\mathrm{we}\mathrm{get}$$$$27\mathrm{y}\mathrm{f}(-\frac{1}{\mathrm{y}})-\frac{1}{{\mathrm{y}}^2}\mathrm{f}\left(\mathrm{y}\right)=-\frac{2}{{\mathrm{y}}^2}...\left(2\right)$$$$\mathrm{Then}27\times \left(1\right)+\mathrm{y}\times \left(2\right)\mathrm{gives}$$$$-\frac{729\mathrm{f}\left(\mathrm{y}\right)}{\mathrm{y}}-\frac{\mathrm{f}\left(\mathrm{y}\right)}{\mathrm{y}}=-{54\mathrm{y}}^2-\frac{2}{\mathrm{y}}$$$$\mathrm{solving}\mathrm{for}\mathrm{f}\left(\mathrm{y}\right),\mathrm{we}\mathrm{have}\mathrm{f}\left(\mathrm{y}\right)=\frac{1}{365}({27\mathrm{y}}^3+1)$$$$\mathrm{or}\mathrm{f}\left(\mathrm{x}\right)=\frac{{27\mathrm{x}}^3+1}{365},\mathrm{thus}\mathrm{f}\left(3\right)=\frac{3^6+1}{365}=2.$$

Commented by bemath last updated on 03/Oct/20

$$\mathrm{santuyy}$$

Answered by mathmax by abdo last updated on 03/Oct/20

$$\mathrm{change}\mathrm{x}\mathrm{by}\frac{1}{\mathrm{x}}\mathrm{we}\mathrm{get}27\mathrm{xf}(-\frac{1}{\mathrm{x}})-\frac{1}{{\mathrm{x}}^2}\mathrm{f}\left(\mathrm{x}\right)=-\frac{2}{{\mathrm{x}}^2}$$$$\mathrm{change}\mathrm{x}\mathrm{by}-\mathrm{x}\mathrm{weget}-27\mathrm{xf}\left(\frac{1}{\mathrm{x}}\right)-\frac{1}{{\mathrm{x}}^2}\mathrm{f}(-\mathrm{x})=-\frac{2}{{\mathrm{x}}^2}\mathrm{so}$$$$\{\begin{array}{l}\frac{27}{\mathrm{x}}\mathrm{f}(-\mathrm{x})-{\mathrm{x}}^2\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)=-{2\mathrm{x}}^2\\ \frac{1}{{\mathrm{x}}^2}\mathrm{f}(-\mathrm{x})+27\mathrm{xf}\left(\frac{1}{\mathrm{x}}\right)=\frac{2}{{\mathrm{x}}^2}\end{array}$$$${\mathrm{\Delta }}_{\mathrm{s}}=\left|\begin{array}{c}\frac{27}{\mathrm{x}}-{\mathrm{x}}^2\\ \frac{1}{{\mathrm{x}}^2}27\mathrm{x}\end{array}\right|={27}^2+1\ne \mathrm{o}\Rightarrow$$$$\mathrm{f}(-\mathrm{x})=\frac{\left|\begin{array}{c}-{2\mathrm{x}}^2-{\mathrm{x}}^2\\ \frac{2}{{\mathrm{x}}^2}27\mathrm{x}\end{array}\right|}{{27}^2+1}=\frac{-{54\mathrm{x}}^3+2}{{27}^2+1}\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{{54\mathrm{x}}^3+2}{{27}^2+1}\Rightarrow \mathrm{f}\left(3\right)=\frac{54\times 3^3+2}{{27}^2+1}$$

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