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Question Number 155973 by Fareed last updated on 06/Oct/21

$$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^3-{3\mathrm{x}}^2+4\mathrm{x}-1$$$$\mathrm{find}\mathrm{a}=?$$$$\mathrm{whenever}\mathrm{f}\left(\mathrm{a}\right)={\mathrm{f}}^{-1}\left(\mathrm{a}\right)$$$$$$

Commented by MJS_new last updated on 06/Oct/21

$$\mathrm{the}\mathrm{graph}\mathrm{of}y=f^{-1}\left(x\right)\mathrm{is}\mathrm{the}\mathrm{mirrored}\mathrm{graph}$$$$\mathrm{of}y=f\left(x\right)\mathrm{for}\mathrm{any}\mathrm{function}.\mathrm{any}\mathrm{intersection}$$$$\mathrm{of}f\left(x\right)=f^{-1}\left(x\right)\mathrm{is}\mathrm{only}\mathrm{possible}\mathrm{oon}\mathrm{the}\mathrm{axis}$$$$\mathrm{of}\mathrm{symmetry}\mathrm{which}\mathrm{is}y=x.$$$$y=f\left(x\right)\wedge y=x\Rightarrow x=f\left(x\right)$$$$\mathrm{we}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{need}{f}^{-1}\left(x\right)$$

Commented by Fareed last updated on 07/Oct/21

$$\mathrm{thanks}$$

Answered by mr W last updated on 06/Oct/21

$${\mathrm{x}}^3-{3\mathrm{x}}^2+4\mathrm{x}-1=x$$$${\mathrm{x}}^3-{3\mathrm{x}}^2+3\mathrm{x}-1=0$$$${(x-1)}^3=0$$$$\Rightarrow x=1=a$$$$i.e.f\left(1\right)=f^{-1}\left(1\right)$$

Commented by Fareed last updated on 06/Oct/21

$$\mathrm{can}\mathrm{you}\mathrm{explian}\mathrm{a}\mathrm{little}\mathrm{more}\mathrm{pleae}$$

Commented by Fareed last updated on 06/Oct/21

$${\mathrm{f}}^{-1}\left(\mathrm{x}\right)=???$$

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