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Summarize the pertinent information obtained by applying the graphing sirategy and sketch the graph of \( y=f(x) \). \( f(x)= \) In \( (x+3)-2 \) Select the correct choice below and, it necessary, fill in the answer box to complete your choice. A. The function f has a local maximum at \( \mathrm{x}= \) (Type an exact answer. Use a comma to separate answers as needed.) Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The function f has a local minimum at \( \mathrm{x}= \) (Type an exact answer. Use a comma to separate answers as needed.) B. The function f has no local minimum Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The function f is concave upward on the subinterval(s) (Type your answer in interval notation. Type an exact answer. Use a comma to separate answers as needed.) B. The function f is never concave upwerd.

Ask by Gibson Ruiz. in the United States
Mar 11,2025

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Answer

The function \( f(x) = \ln(x+3) - 2 \) is strictly increasing with no local maxima or minima. It is concave downward on its entire domain \( x > -3 \) and has no intervals where it is concave upward.

Solution

1. **Determine the Domain and Behavior** The function is \[ f(x)=\ln(x+3)-2. \] The domain is determined by the argument of the logarithm: \[ x+3>0 \quad\Longrightarrow\quad x>-3. \] Thus, the graph is defined for \( x>-3 \) with a vertical asymptote at \( x=-3 \). 2. **Identify Monotonicity** Compute the first derivative: \[ f'(x)=\frac{1}{x+3}. \] Since \( x+3>0 \) for \( x>-3 \), we have \( f'(x)>0 \). Therefore, the function is strictly increasing on its domain. **Implication:** - There is no local maximum because the function never decreases. - There is no local minimum because there is no change from decreasing to increasing within the domain (the endpoint \( x=-3 \) is not included in the domain). 3. **Determine Concavity** Compute the second derivative: \[ f''(x)=-\frac{1}{(x+3)^2}. \] Since \((x+3)^2>0\) for all \(x>-3\), it follows that \( f''(x)<0 \) for all \( x>-3 \). **Implication:** - The graph is concave downward on its entire domain. - There is no interval of concave upward. 4. **Graph Sketch Summary** - **Domain:** \(x>-3\) with a vertical asymptote at \(x=-3\). - **Intercept:** When \(x= -2\), \(f(-2)=\ln(1)-2=0-2=-2\). (More intercepts can be found similarly if needed.) - **Monotonicity:** The function is strictly increasing. - **Concavity:** The function is concave downward. 5. **Answering the Multiple-Choice Selections** - Local maximum: Since the function is strictly increasing, there is no local maximum. - **Choice:** (No local maximum exists.) - Local minimum: Similarly, no local minimum exists. - **Choice:** \(B.\) The function \(f\) has no local minimum. - Concavity: The function is never concave upward. - **Choice:** \(B.\) The function \(f\) is never concave upward.

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To analyze the function \( f(x) = \ln(x + 3) - 2 \), we first identify its domain, which is \( x > -3 \). The function is defined for values greater than -3, and the natural logarithm will ensure that \( f(x) \) approaches negative infinity as \( x \) approaches -3 from the right. By finding the derivative, \( f'(x) = \frac{1}{x + 3} \), we see that \( f'(x) > 0 \) for \( x > -3 \), indicating that the function is strictly increasing in this interval. Therefore, there are no local maximums or minimums. Since \( f'(x) > 0 \) for all \( x > -3 \) and \( f''(x) = -\frac{1}{(x + 3)^2} < 0 \) indicates it is concave downward everywhere on its domain, we can deduce that it is never concave upward. A. The function f has a local maximum at \( \mathrm{x}= \) None, A. The function f has a local minimum at \( \mathrm{x}= \) None, B. The function f has no local minimum. A. The function f is concave upward on the subinterval(s) None, B. The function f is never concave upward.

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