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

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(B) Given that $f^{\prime}(x)=4 x^3-3 x^{-4}$.Integrating both sides with respect to $x$:$f(x) = \int (4x^3 - 3x^{-4}) dx$$f(x) = 4 \cdot \frac{x^4}{4

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given that $f^{\prime}(x)=4 x^3-3 x^{-4}$.Integrating both sides with respect to $x$:$f(x) = \int (4x^3 - 3x^{-4}) dx$$f(x) = 4 \cdot \frac{x^4}{4} - 3 \cdot \frac{x^{-3}}{-3} + c$$f(x) = x^4 + \frac{1}{x^3} + c$Given that $f(2) = 0$,substitute $x = 2$:$0 = (2)^4 + \frac{1}{2^3} + c$$0 = 16 + \frac{1}{8} + c$$0 = \frac{128+1}{8} + c$$c = -\frac{129}{8}$Therefore,$f(x) = x^4 + \frac{1}{x^3} - \frac{129}{8}$.

If $\frac{d}{d x} f(x)=4 x^3-\frac{3}{x^4}$ such that $f(2)=0$,then $f(x)$ is equal to

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