Answer
False
The most general antiderivative of $f(x) = x^{-2}$ is:
$F(x) = -\frac{1}{x}+C_1$ if $x\lt 0$
$F(x) = -\frac{1}{x}+C_2$ if $x\gt 0$
Work Step by Step
We can verify the derivative of $F(x)$:
$F(x) = -\frac{1}{x}+C$
$F(x) = -x^{-1}+C$
$F'(x) = x^{-2}$
$F'(x) = f(x)$ so $F(x)$ is an anti-derivative of $f(x)$.
However, consider that the function is not defined at $x=0$. Thus, the most general anti-derivative should be written as a piecewise function excluding $x=0$ (see Example 4.9#1b).
$F(x) = -\frac{1}{x}+C_1$ if $x\lt 0$
$F(x) = -\frac{1}{x}+C_2$ if $x\gt 0$
Thus the statement is technically False.
