St I
x < -1
if x = -2 we have \((-2)^3+9\) = 1 f(x) is positive.
if x = -3 we have \((-3)^3+9\) = -18 f(x) is negative ----------Insufficient.

St II
x > -3
if x = -2 we have \((-2)^3+9\) = 1 f(x) is positive.
if x = -2.5 we have \((-2.5)^3+9\) = -6.25 f(x) is negative. ----------Insufficient.

St. I and II ----------Insufficient.

Hence Option E is correct.
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Official solution from Veritas Prep.

If \(x<−1\), then x could be \(−2\) in which case \(f(x)\) is positive or \(x\) could be \(−3\) in which case \(f(x)\) is negative. Therefore, Statement 1 alone is not sufficient.

If \(x>−3\), the x could be \(−2\) in which case \(f(x)\) is positive. However, the problem never stated that \(x\) is an integer. Then, \(x\) could be \(-2.99999\), that is almost \(-3\). In this case, \(f(x)\) will be a negative. \(f(x)=(-2.99999)^3+9 = -17.99973\) Therefore, Statement 2 is not sufficient.

Combining Statement 1 and Statement 2, \(f(x)\) could still either be positive or negative. Therefore, the answer is \(E\).
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(-2.9999)^3 = -26.997300 < 0.