A differentiable function satisfies 3f^2(x) f | vedclass

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(A) Given the differential equation: $3f^2(x) f'(x) = 2x$. Integrating both sides with respect to $x$: $\int 3f^2(x) f'(x) dx = \int 2x dx$. Using the

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$\sqrt[3]{6}$

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(A) Given the differential equation: $3f^2(x) f'(x) = 2x$. Integrating both sides with respect to $x$: $\int 3f^2(x) f'(x) dx = \int 2x dx$. Using the substitution $u = f(x)$,$du = f'(x) dx$,we get: $\int 3u^2 du = x^2 + C$. $u^3 = x^2 + C$,which means $f^3(x) = x^2 + C$. Given $f(2) = 1$,substitute $x = 2$ and $f(2) = 1$: $1^3 = 2^2 + C 1 = 4 + C C = -3$. So,the function is $f^3(x) = x^2 - 3$. To find $f(3)$,substitute $x = 3$: $f^3(3) = 3^2 - 3 = 9 - 3 = 6$. Therefore,$f(3) = \sqrt[3]{6}$.

$A$ differentiable function satisfies $3f^2(x) f'(x) = 2x$. Given $f(2) = 1$,then the value of $f(3)$ is:

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