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Question Number 172912 by mnjuly1970 last updated on 03/Jul/22

Answered by FongXD last updated on 03/Jul/22

$$★\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}(\mathrm{x}+4\mathrm{n})$$$$\Rightarrow {\mathrm{f}}^{\prime }\left(\mathrm{x}\right)={\mathrm{f}}^{\prime }(\mathrm{x}+4\mathrm{n})$$$$★\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}(\mathrm{x}+1)+\mathrm{f}(3\mathrm{x}+2)$$$$\Rightarrow {\mathrm{f}}^{\prime }\left(\mathrm{x}\right)={\mathrm{f}}^{\prime }(\mathrm{x}+1)+{3\mathrm{f}}^{\prime }(3\mathrm{x}+2)$$$$\bullet \mathrm{x}=1,{\mathrm{f}}^{\prime }\left(1\right)={\mathrm{f}}^{\prime }\left(2\right)+{3\mathrm{f}}^{\prime }\left(5\right)={\mathrm{f}}^{\prime }\left(2\right)+{3\mathrm{f}}^{\prime }\left(1\right)$$$$\Rightarrow {\mathrm{f}}^{\prime }\left(2\right)=-{2\mathrm{f}}^{\prime }\left(1\right)$$$$\bullet \mathrm{x}=0,{\mathrm{f}}^{\prime }\left(0\right)={\mathrm{f}}^{\prime }\left(1\right)+{3\mathrm{f}}^{\prime }\left(2\right)={\mathrm{f}}^{\prime }\left(1\right)-{6\mathrm{f}}^{\prime }\left(1\right)$$$$\Rightarrow {\mathrm{f}}^{\prime }\left(0\right)=-{5\mathrm{f}}^{\prime }\left(1\right)$$$$\bullet \mathrm{x}=2,{\mathrm{f}}^{\prime }\left(2\right)={\mathrm{f}}^{\prime }\left(3\right)+{3\mathrm{f}}^{\prime }\left(8\right)=-18+{3\mathrm{f}}^{\prime }\left(0\right)$$$$\iff -{2\mathrm{f}}^{\prime }\left(1\right)=-18-{15\mathrm{f}}^{\prime }\left(1\right)$$$$\Rightarrow {\mathrm{f}}^{\prime }\left(1\right)=-\frac{18}{13}$$

Commented by mnjuly1970 last updated on 03/Jul/22

$$\mathrm{thanks}\mathrm{alot}$$

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