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

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Aicharya · Accepted Answer

Given that $\mathrm{f}^{\prime}(x)=4 x^3-3 x^{-4}$ $\begin{array}{ll} \therefore & \mathrm{f}(x)=\int\left(4 x^3-3 x^{-4}\right) \mathrm{d} x \\ \therefore & \mathrm{f}(x)=x^4+\frac{1}{x^3}+\mathrm{c} \end{array}$
Given that $\mathrm{f}(2)=0$ $\begin{array}{ll} \therefore & 16+\frac{1}{8}+c=0 \\ \therefore & c=\frac{-129}{8} \\ \therefore & \mathrm{f}(x)=x^4+\frac{1}{x^3}-\frac{129}{8} \end{array}$

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

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