\[ R'(x) = \frac{d}{dx}[\ln(x^2 + 4x + 5)] = \frac{2x + 4}{x^2 + 4x + 5} \]
![\[ R'(x) = \frac{d}{dx}[\ln(x^2 + 4x + 5)] = \frac{2x + 4}{x^2 + 4x + 5} \]](https://soloferat.biz.id/images/rx--fracddxlnx2--4x--5--frac2x--4x2--4x--5-.jpg)
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- A health data analyst models patient recovery rates with the function \( R(x) = \ln(x^2 + 4x + 5) \). Find the value of \( x \) that maximizes \( R(x) \).
- To maximize \( R(x) = \ln(x^2 + 4x + 5) \), we first find the derivative \( R'(x) \) and set it to zero.
- Differentiate \( R(x) \):
- Set the derivative to zero:
- \[ \frac{2x + 4}{x^2 + 4x + 5} = 0 \]
- The numerator must be zero, so:
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