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Summarize the pettinent information ottained by applying the graphing stralegy and sketch the graph of \( y=f(x) \). \( f(x)=\operatorname{In}(x+8)-1 \) Select the correct choice below and, if necessary, fill in the answer box to complele your choice. A. The function \( f \) is increasing on the subinterval(s) ( \( -8, \infty \) ) (Type your answer in interval notation. Use a comma to separate answers as needed.) Belect the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The function \( f \) is decreasing on the subinterval(s) (Type your answer in interval notation. Use a comma to separate answers as needed.) B. The function \( f \) is never decreasing. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The function \( f \) has a local maximum at \( x=\square \). (Type an exact answer Use a comma to separale answers as needed.) B. The function \( f \) has no local maximum.

Ask by John Pena. in the United States
Mar 11,2025

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Answer

The function \( f(x) = \ln(x+8) - 1 \) is strictly increasing on the interval \( (-8, \infty) \) and never decreasing. It has no local maximum.

Solution

**Step 1. Determine the Domain** The function is \[ f(x)=\ln(x+8)-1. \] Since the logarithm is defined only when its argument is positive, we require \[ x+8>0 \quad \Longrightarrow \quad x>-8. \] Thus, the domain is \[ (-8,\infty). \] **Step 2. Monotonicity (Increasing/Decreasing Behavior)** Differentiate \( f(x) \): \[ f'(x)=\frac{d}{dx} \left[\ln(x+8)-1\right]=\frac{1}{x+8}. \] Since \( x+8>0 \) for all \( x\in(-8,\infty) \), it follows that \[ f'(x)>0 \quad \text{for all} \quad x\in(-8,\infty). \] Therefore, the function is strictly increasing on its entire domain. **Step 3. Local Extrema** Because \( f \) is strictly increasing on \( (-8,\infty) \), it has no turning points. In particular, there is no local maximum. **Step 4. Other Graph Details** - **Vertical Asymptote:** The domain restriction \( x>-8 \) indicates a vertical asymptote at \[ x=-8. \] - **\( x \)-Intercept:** Solve \( f(x)=0 \): \[ \ln(x+8)-1=0 \quad \Longrightarrow \quad \ln(x+8)=1 \quad \Longrightarrow \quad x+8=e \quad \Longrightarrow \quad x=e-8. \] So, the graph crosses the \( x \)-axis at \[ x=e-8. \] - **\( y \)-Intercept:** Evaluate \( f(0) \): \[ f(0)=\ln(8)-1. \] **Step 5. Sketching the Graph** - Draw a vertical dashed line at \( x=-8 \) to represent the vertical asymptote. - Plot the \( x \)-intercept at \( \left(e-8,\,0\right) \) and the \( y \)-intercept at \( \left(0,\,\ln(8)-1\right) \). - The graph is increasing on \( (-8,\infty) \) and approaches \( -\infty \) as \( x \) approaches \( -8 \) from the right. - As \( x\to \infty \), \( f(x) \) increases without bound (though slowly due to the logarithm). **Answering the Multiple-Choice Selections** 1. **Increasing Intervals:** The function is increasing on the subinterval(s) \[ (-8,\infty). \] 2. **Decreasing Intervals:** Since \( f'(x)>0 \) for all \( x\in(-8,\infty) \), the function is never decreasing. 3. **Local Maximum:** The function is strictly increasing and has no turning point, so it has no local maximum. Thus, the correct choices are: - For the increasing interval: \[ (-8,\infty) \] - For the decreasing interval: The function is never decreasing. - For the local maximum: The function has no local maximum.

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To analyze the function \( f(x) = \ln(x + 8) - 1 \), we first note that the natural logarithm function is defined for \( x + 8 > 0 \), meaning \( x > -8 \). The derivative of the function is \( f'(x) = \frac{1}{x + 8} \), which is always positive for \( x > -8 \). This indicates that the function is increasing over the interval \( (-8, \infty) \). Thus, the correct responses are: A. The function \( f \) is increasing on the subinterval(s) ( -8, ∞ ). Since the derivative \( f'(x) \) does not equal zero or change sign in this interval, we can conclude: B. The function \( f \) has no local maximum.

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