If frac{d}{d x}(f(x))=4 x^3-frac{3}{x^4} and | vedclass

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(A) Given that $\frac{d}{d x}(f(x))=4 x^3-3 x^{-4}$.To find $f(x)$,we integrate both sides with respect to $x$:$f(x) = \int (4 x^3-3 x^{-4}) d x$.$f(x

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$x^4+\frac{1}{x^3}-\frac{129}{8}$

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(A) Given that $\frac{d}{d x}(f(x))=4 x^3-3 x^{-4}$.To find $f(x)$,we integrate both sides with respect to $x$:$f(x) = \int (4 x^3-3 x^{-4}) d x$.$f(x) = 4 \frac{x^4}{4} - 3 \frac{x^{-3}}{-3} + C$.$f(x) = x^4 + x^{-3} + C = x^4 + \frac{1}{x^3} + C$.Given $f(2) = 0$,substitute $x=2$:$f(2) = 2^4 + \frac{1}{2^3} + C = 0$.$16 + \frac{1}{8} + C = 0$.$C = - (16 + \frac{1}{8}) = - \frac{128+1}{8} = - \frac{129}{8}$.Therefore,$f(x) = x^4 + \frac{1}{x^3} - \frac{129}{8}$.

If $\frac{d}{d x}(f(x))=4 x^3-\frac{3}{x^4}$ and $f(2)=0$,then $f(x)=$ . . . . . . .

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