If $f^{'} (x) = \frac{2}{x^{2}}$ then...
A. $f (x) = - 2 x^{- 1} + C$
B. $f (x) = - 4 x^{- 3} + C$
C. $f (x) = - 2 x^{- 3} + C$
D. $f (x) = 2 x^{- 1} + C$

Answer

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Final Answer: The correct answer is $f (x) = - 2 x^{- 1} + C$ , which corresponds to option A.

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Step 1 ：The question is asking for the antiderivative of the function $f^{'} (x) = \frac{2}{x^{2}}$ . The antiderivative of a function is the function whose derivative is the given function. In this case, we need to find the function $f (x)$ such that its derivative is $f^{'} (x) = \frac{2}{x^{2}}$ .

Step 2 ：The antiderivative of $f^{'} (x) = \frac{2}{x^{2}}$ is $- 2 / x$ .

Step 3 ：This means that the function $f (x)$ is $- 2 / x$ plus a constant $C$ .

Step 4 ：Final Answer: The correct answer is $f (x) = - 2 x^{- 1} + C$ , which corresponds to option A.

If f′(x)=2x2 then...A. f(x)=−2x−1+CB. f(x)=−4x−3+CC. f(x)=−2x−3+CD. f(x)=2x−1+C

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If $f^{'} (x) = \frac{2}{x^{2}}$ then...
A. $f (x) = - 2 x^{- 1} + C$
B. $f (x) = - 4 x^{- 3} + C$
C. $f (x) = - 2 x^{- 3} + C$
D. $f (x) = 2 x^{- 1} + C$