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Question Number 106309 by bemath last updated on 04/Aug/20

$$\mathrm{If}\mathrm{g}\left(\mathrm{x}\right)=\mathrm{x}+\sqrt{\mathrm{x}}\mathrm{and}\underset{x\to 2}{\lim }\frac{\mathrm{f}\left(\mathrm{x}\right)-\mathrm{f}\left(2\right)}{{\mathrm{x}}^2+\mathrm{ax}+\mathrm{b}}=\frac{4}{3}$$$$\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}{(\mathrm{f}\circ \mathrm{g})}^{\prime }\left(1\right).$$

Answered by bobhans last updated on 05/Aug/20

$$\mathrm{g}\left(\mathrm{x}\right)=\mathrm{x}+\sqrt{\mathrm{x}}\Rightarrow {\mathrm{g}}^{\prime }\left(\mathrm{x}\right)=1+\frac{1}{2\sqrt{\mathrm{x}}};{\mathrm{g}}^{\prime }\left(1\right)=\frac{3}{2}$$$$\mathrm{g}\left(1\right)=2$$$${(\mathrm{f}\circ \mathrm{g})}^{\prime }\left(1\right)={\mathrm{g}}^{\prime }\left(1\right)\mathrm{f}{}^{\prime }\left(\mathrm{g}\left(1\right)\right)=\frac{3}{2}\mathrm{f}{}^{\prime }\left(2\right)$$$$\left(1\right)2^2+2\mathrm{a}+\mathrm{b}=0\to \mathrm{b}=-4-2\mathrm{a}$$$$\left(2\right)\underset{x\to 2}{\lim }\frac{\mathrm{f}\left(\mathrm{x}\right)-\mathrm{f}\left(2\right)}{{\mathrm{x}}^2+\mathrm{ax}-2\mathrm{a}-4}=\frac{4}{3}$$$$\underset{x\to 2}{\lim }\frac{\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)}{2\mathrm{x}+\mathrm{a}}=\frac{4}{3}$$$$\mathrm{case}\left(1\right)\mathrm{if}\mathrm{a}\ne -4,\mathrm{then}\mathrm{we}\mathrm{get}$$$$\mathrm{f}{}^{\prime }\left(2\right)=\frac{16+4\mathrm{a}}{3}.\mathrm{so}{(\mathrm{f}\circ \mathrm{g})}^{\prime }\left(1\right)={\mathrm{g}}^{\prime }\left(1\right).\mathrm{f}{}^{\prime }\left(\mathrm{g}\left(1\right)\right)$$$$=\frac{3}{2}.\mathrm{f}{}^{\prime }\left(2\right)=\frac{3}{2}.\frac{16+4\mathrm{a}}{3}=8+2\mathrm{a}$$$$\mathrm{case}\left(2\right)\mathrm{if}\mathrm{a}=-4,\mathrm{then}\underset{x\to 2}{\lim }\frac{\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)}{2\mathrm{x}+\mathrm{a}}\mathrm{is}\mathrm{form}\frac{0}{0}$$$$\mathrm{so}\mathrm{f}{}^{\prime }\left(2\right)=0\Rightarrow {(\mathrm{f}\circ \mathrm{g})}^{\prime }\left(1\right)=\frac{3}{2}.0=0$$$$$$$$$$

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