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Question Number 118663 by benjo_mathlover last updated on 19/Oct/20

$$Givenf\left(x\right)=\int \underset{0}{\stackrel{x}{}}\frac{dt}{{\left[f\left(t\right)\right]}^2}and\int \underset{0}{\stackrel{2}{}}\frac{dt}{{\left[f\left(t\right)\right]}^2}=\sqrt[3]{6}$$$$Thenthenthevalueoff\left(9\right)is\mathrm{\_}\mathrm{\_}$$

Commented by mr W last updated on 19/Oct/20

$$f{}^{\prime }\left(x\right)=\frac{1}{{\left(f\left(x\right)\right)}^2}$$$$y^{2}dy=dx$$$$\frac{y^3}{3}=x+\frac{C}{3}$$$$y^3=3x+C$$$${\left(\sqrt[3]{6}\right)}^3=3\times 2+C$$$$\Rightarrow C=0$$$$\Rightarrow y^3=3x$$$$\Rightarrow y=f\left(x\right)=\sqrt[3]{3x}$$$$f\left(9\right)=\sqrt[3]{3\times 9}=3$$

Commented by 1549442205PVT last updated on 19/Oct/20

$$\mathrm{Great}!\mathrm{Sir}.$$

Answered by benjo_mathlover last updated on 19/Oct/20

$$byTheoremFundamentalCalculus-1$$$$f{}^{\prime }\left(x\right)=\underset{0}{\stackrel{x}{\int }}f\left(t\right)dt.Givenequation$$$$f\left(x\right)=\underset{0}{\stackrel{x}{\int }}\frac{dt}{{\left[f\left(t\right)\right]}^2}\Rightarrow f{}^{\prime }\left(x\right)=\frac{1}{{\left[f\left(x\right)\right]}^2}$$$$\Rightarrow {\left[f\left(x\right)\right]}^{2}d\left(f\left(x\right)\right)=dx,integrating$$$$bothsides\int {\left[f\left(x\right)\right]}^{2}dx=\int dx$$$$\frac{1}{3}{\left[f\left(x\right)\right]}^3=x+C$$$$\Rightarrow sof\left(x\right)=\sqrt[3]{3x+\lambda },\lambda =3C$$$$thus{\int }_0^2\frac{dt}{{\left[f\left(t\right)\right]}^2}=f\left(2\right)=\sqrt[3]{6}$$$$wegetf\left(2\right)=\sqrt[3]{6+\lambda }=\sqrt[3]{6},give\lambda =0$$$$Thusf\left(x\right)=\sqrt[3]{3x}andf\left(9\right)=\sqrt[3]{27}=3$$

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