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Question Number 144629 by mathdanisur last updated on 27/Jun/21

$$if{f}^2(2x-1)-10f(3x-2)+25=0$$$$findf{}^{{}^{\prime }}\left(1\right)+f\left(1\right)=?$$

Answered by liberty last updated on 27/Jun/21

$${\left[\mathrm{f}(2\mathrm{x}-1)\right]}^2-10\mathrm{f}(3\mathrm{x}-2)+25=0$$$$\frac{\mathrm{d}}{\mathrm{dx}}[{\left(\mathrm{f}(2\mathrm{x}-1)\right)}^2-10\mathrm{f}(3\mathrm{x}-2)+25]=0$$$$\Rightarrow 2.2\mathrm{f}(2\mathrm{x}-1)\mathrm{f}{}^{\prime }(2\mathrm{x}-1)-30\mathrm{f}{}^{\prime }(3\mathrm{x}-2)=0$$$$\mathrm{case}\left(1\right)2\mathrm{x}-1=1;\mathrm{x}=1$$$$\to 4\mathrm{f}\left(1\right).\mathrm{f}{}^{\prime }\left(1\right)-30\mathrm{f}{}^{\prime }\left(1\right)=0$$$$\Rightarrow \mathrm{f}{}^{\prime }\left(1\right)\{4\mathrm{f}\left(1\right)-30\}=0$$$$\{\begin{array}{l}\mathrm{f}{}^{\prime }\left(1\right)=0\\ \mathrm{f}\left(1\right)=\frac{15}{2}\left(\mathrm{rejected}\right)\end{array}$$$$\mathrm{case}\left(2\right)\mathrm{inserting}\mathrm{x}=1\mathrm{to}\mathrm{original}$$$$\mathrm{equation}$$$$\Rightarrow {\left[\mathrm{f}\left(1\right)\right]}^2-10\mathrm{f}\left(1\right)+25=0$$$$\Rightarrow {(\mathrm{f}\left(1\right)-5)}^2=0\Rightarrow \mathrm{f}\left(1\right)=5$$$$\mathrm{therefore}\mathrm{f}{}^{\prime }\left(1\right)+\mathrm{f}\left(1\right)=0+5=5$$

Commented by mathdanisur last updated on 27/Jun/21

$$alotcoolthankyouSir$$

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