What is the range of (f(x)=-\frac{3}{(x+1)^2}+2)?

Correct answer: ((-\infty,2))

The direct answer is option A: \((-\infty,2)\). The function is \(f(x)=-\frac{3}{(x+1)^2}+2\). First, x cannot equal −1 because the denominator would be zero. For every allowed x, \((x+1)^2>0\). Dividing 3 by this positive number gives a positive value, and the negative sign makes \(-\frac{3}{(x+1)^2}<0\). Therefore f(x)=2−a for some a>0, so f(x)<2. The value 2 is never reached because the fraction can never be zero. As x becomes very large in either direction, the fraction approaches zero and f(x) approaches 2 from below, but never equals it. Hence the range is \((-\infty,2)\). Option A is correct. Option B, \((2,\infty)\), is wrong because every output is below 2. Option C, \((-\infty,2]\), incorrectly includes 2. Option D, \([2,\infty)\), has the wrong side and incorrectly includes 2. The domain restriction x≠−1 does not create a missing output other than the upper endpoint; all values less than 2 can occur. Remember: a negative positive fraction added to 2 gives values strictly less than 2.