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Question Number 145609 by Khalmohmmad last updated on 06/Jul/21

Commented by imjagoll last updated on 06/Jul/21

$$\mathrm{let}\mathrm{g}\left(\mathrm{x}\right)={\mathrm{f}}^{-1}\left(\mathrm{x}\right)$$$$\Rightarrow (\mathrm{g}\bullet \mathrm{f})\left(\mathrm{x}\right)=\mathrm{x}$$$$\Rightarrow \mathrm{f}{}^{\prime }\left(\mathrm{x}\right).{\mathrm{g}}^{\prime }\left(\mathrm{f}\left(\mathrm{x}\right)\right)=1$$$$\Rightarrow \mathrm{g}{}^{\prime }\left(\mathrm{f}\left(\mathrm{x}\right)\right)=\frac{1}{\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)}=\frac{1}{{8\mathrm{x}}^3+{12\mathrm{x}}^2+8}$$$$\Rightarrow {\left[{\mathrm{f}}^{-1}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\right]}^{\prime }=\frac{1}{{8\mathrm{x}}^3+{12\mathrm{x}}^2+8}$$$$\mathrm{f}\left(\mathrm{x}\right)=-3\Rightarrow {2\mathrm{x}}^4+{4\mathrm{x}}^3+8\mathrm{x}+7=-3$$$$\Rightarrow {2\mathrm{x}}^4+{4\mathrm{x}}^3+8\mathrm{x}+10=0$$$$\Rightarrow {\mathrm{x}}^4+{2\mathrm{x}}^3+4\mathrm{x}+5=0$$$$\Rightarrow (\mathrm{x}+1)({\mathrm{x}}^3+{\mathrm{x}}^2-\mathrm{x}+5)=0$$$$\Rightarrow \mathrm{x}=-1$$$$\mathrm{thus}{\left[{\mathrm{f}}^{-1}(-3)\right]}^{\prime }=\frac{1}{8(-1)+12\left(1\right)+8}$$$$=\frac{1}{12}$$

Answered by gsk2684 last updated on 06/Jul/21

$$f\left(x\right)=2x^4+4x^3+8x+7$$$$f\left(x\right)=-3\Rightarrow 2x^4+4x^3+8x+10=0$$$$\Rightarrow 2(x+1)(x^3+x^2-x+5)=0$$$$f^{-1}(-3)=-1$$$$and{f}^{\mid }\left(x\right)=8x^3+12x^2+8$$$$f\left(f^{-1}\left(x\right)\right)=x$$$$f^{}\left(f^{-1}\left(x\right)\right){\left(f^{-1}\left(x\right)\right)}^{}=1$$$${\left(f^{-1}\left(x\right)\right)}^{}=\frac{1}{f^{}\left(f^{-1}\left(x\right)\right)}$$$${\left(f^{-1}(-3)\right)}^{}=\frac{1}{f^{}\left(f^{-1}(-3)\right)}=\frac{1}{f^{}(-1)}$$$$$$

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