Let \( f(x)=\ln \left(e^{x}-9\right) \)

The function you provided is \[ f(x) = \ln(e^x - 9). \] To analyze this function, we can consider its domain, range, and some properties. ### Domain The argument of the logarithm must be positive, so we need: \[ e^x - 9 > 0. \] This simplifies to: \[ e^x > 9. \] Taking the natural logarithm of both sides gives: \[ x > \ln(9). \] Thus, the domain of \( f(x) \) is: \[ x \in (\ln(9), \infty). \] ### Range As \( x \) approaches \( \ln(9) \) from the right, \( e^x \) approaches \( 9 \), and thus \( f(x) \) approaches \( \ln(0) \), which tends to \( -\infty \). As \( x \) increases without bound, \( e^x \) also increases without bound, and so \( f(x) \) approaches \( \ln(\infty) \), which tends to \( \infty \). Therefore, the range of \( f(x) \) is: \[ (-\infty, \infty). \] ### Derivative To find the derivative of \( f(x) \), we can use the chain rule. The derivative of \( \ln(u) \) is \( \frac{1}{u} \cdot u' \). Here, \( u = e^x - 9 \), so: \[ u' = e^x. \] Thus, the derivative \( f'(x) \) is: \[ f'(x) = \frac{1}{e^x - 9} \cdot e^x = \frac{e^x}{e^x - 9}. \] ### Summary - The function is defined for \( x > \ln(9) \). - The range of \( f(x) \) is \( (-\infty, \infty) \). - The derivative is given by: \[ f'(x) = \frac{e^x}{e^x - 9}. \] If you have any specific questions or need further analysis, feel free to ask!