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

 f(x)=∫_1 ^x  (dt/( (√(t^3 +2t^2 +3))))     (f^(−1) (0))′=?

$$\:\mathrm{f}\left(\mathrm{x}\right)=\int_{\mathrm{1}} ^{{x}} \:\frac{{dt}}{\:\sqrt{{t}^{\mathrm{3}} +\mathrm{2}{t}^{\mathrm{2}} +\mathrm{3}}} \\ $$$$\:\:\:\left({f}^{−\mathrm{1}} \left(\mathrm{0}\right)\right)'=? \\ $$

Commented by cortano1 last updated on 18/Feb/22

 not (√6) ?

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

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

(f^(−1) (0))′=(1/(f′(f^(−1) (0))))=(1/(f^′ (1)))=((1/( (√(1^3 +2×1^2 +3)))))^(−1) =(√6)

$$\left(\mathrm{f}^{−\mathrm{1}} \left(\mathrm{0}\right)\right)'=\frac{\mathrm{1}}{\mathrm{f}'\left(\mathrm{f}^{−\mathrm{1}} \left(\mathrm{0}\right)\right)}=\frac{\mathrm{1}}{\mathrm{f}^{'} \left(\mathrm{1}\right)}=\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}^{\mathrm{3}} +\mathrm{2}×\mathrm{1}^{\mathrm{2}} +\mathrm{3}}}\right)^{−\mathrm{1}} =\sqrt{\mathrm{6}} \\ $$

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

yes, (√6) is right.

$${yes},\:\sqrt{\mathrm{6}}\:{is}\:{right}. \\ $$

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

(f^(−1) (0))′=(f^(−1) (x))′∣_(x=0)

$$\left(\mathrm{f}^{−\mathrm{1}} \left(\mathrm{0}\right)\right)'=\left(\mathrm{f}^{−\mathrm{1}} \left({x}\right)\right)'\mid_{{x}=\mathrm{0}} \\ $$

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

see Q166331

$${see}\:{Q}\mathrm{166331} \\ $$

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