Question

If $\frac {d}{dx}\ f(x)= 4x^3 - \frac{3}{x^4}$ such that f(2)=0. Then f(x) is
(A)$x^4+\frac {1}{x^{3}}-\frac {129}{8}$
(B)$x^3+\frac {1}{x^{4}}+\frac {129}{8}$
(C)$x^4+\frac {1}{x^{3}}+\frac {129}{8}$
(D)$x^3+\frac {1}{x^{4}}-\frac {129}{8}$

Answer 1

Answer:

A

Step-by-step explanation:

$\displaystyle \frac{df}{dx} = 4x^3 - 3x^{-4}\\ \\ \Rightarrow f = 4\left(\frac{x^4}{4}\right) - 3\left(\frac{x^{-3}}{-3}\right) + C \\ \\= x^4 + \frac{1}{x^3} + C$

From here,

0 = f(2) = 2⁴ + 1/2³ + C = 16 + 1/8 + C = 129/8 + C

=> C = -129/8