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Question Number 166294 by cortano1 last updated on 18/Feb/22

$$\mathrm{f}\left(\mathrm{x}\right)={\int }_1^x\frac{dt}{\sqrt{t^3+2t^2+3}}$$$${\left(f^{-1}\left(0\right)\right)}^{\prime }=?$$

Commented by cortano1 last updated on 18/Feb/22

$$\mathrm{not}\sqrt{6}?$$

Commented by mr W last updated on 18/Feb/22

$${\left({\mathrm{f}}^{-1}\left(0\right)\right)}^{\prime }=\frac{1}{{\mathrm{f}}^{\prime }\left({\mathrm{f}}^{-1}\left(0\right)\right)}=\frac{1}{{\mathrm{f}}^{{}^{\prime }}\left(1\right)}={\left(\frac{1}{\sqrt{1^3+2\times 1^2+3}}\right)}^{-1}=\sqrt{6}$$

Commented by mr W last updated on 18/Feb/22

$$yes,\sqrt{6}isright.$$

Commented by mr W last updated on 18/Feb/22

$${\left({\mathrm{f}}^{-1}\left(0\right)\right)}^{\prime }={\left({\mathrm{f}}^{-1}\left(x\right)\right)}^{\prime }{\mid }_{x=0}$$

Commented by mr W last updated on 18/Feb/22

$$seeQ166331$$

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