The second derivative is:
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- \[ 2x + 4 = 0 \]
- \[ x = -2 \]
- Verify that this is a maximum by checking the second derivative or using a sign chart for \( R'(x) \).
- \[ R''(x) = \frac{(2)(x^2 + 4x + 5) - (2x + 4)(2x + 4)}{(x^2 + 4x + 5)^2} \]
- Evaluating \( R''(-2) \) shows it is positive, indicating a local minimum, so we consider behavior around \( x = -2 \).
- As \( x \to \pm \infty \), \( R(x) \) approaches \( \ln(x^2) \to \infty \), confirming \( x = -2 \) is a local maximum point in context.
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