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Summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of \( \mathrm{f}(\mathrm{x})=5+2 e^{-0.2 \mathrm{x}} \). Find the intervals where \( \mathrm{f}(\mathrm{x}) \) is increasing or decreasing. Select the correct choice below and fill in the answer box(es) to complete your choice. (Type your answer in interval notation. Use a comma to separate answers as needed.) A. The function is increasing on \( \square \). It is never decreasing. B. The function is increasing on \( \square \). It is decreasing on \( \square \). C. The function is decreasing on \( \square \). It is never increasing. Aind the location of any local extrema of \( \mathrm{f}(\mathrm{x}) \). Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice. (Type integers or decimals. Use a comma to separate answers as needed.) B. There is a local minimum at \( \mathrm{x}=\square \). There is no local maximum. (Type an integer or a decimal. Use a comma to separate answers as needed.) C. There is a local maximum at \( \mathrm{x}=\square \). There is no local minimum. (Type an integer or a decimal. Use a comma to separate answers as needed.) D. There are no local extrema.

Ask by John Gordon. in the United States
Mar 17,2025

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Tutor-Verified Answer

Answer

The function \( \mathrm{f}(x) = 5 + 2e^{-0.2x} \) is strictly decreasing on \( (-\infty, \infty) \) and has no local extrema.

Solution

1. Compute the derivative: \[ \mathrm{f}(x) = 5 + 2e^{-0.2x} \] Differentiate using the chain rule: \[ \mathrm{f}'(x) = 2 \cdot (-0.2)e^{-0.2x} = -0.4e^{-0.2x} \] Since \( e^{-0.2x} > 0 \) for all \( x \), we have \[ \mathrm{f}'(x) < 0 \quad \text{for all } x. \] 2. Determine intervals of increase/decrease: Because the derivative is always negative, the function is strictly decreasing on its entire domain, \( (-\infty, \infty) \). 3. Identify the local extrema: Since \( \mathrm{f}(x) \) is strictly decreasing, there is no point at which the derivative changes sign. Therefore, the function has no local maximum or local minimum. 4. Graph summary: - **Y-intercept:** When \( x=0 \), \[ \mathrm{f}(0) = 5 + 2e^{0} = 5 + 2 = 7. \] - **Horizontal asymptote:** As \( x \to \infty \), \( e^{-0.2x} \to 0 \) so \[ \mathrm{f}(x) \to 5. \] - **End behavior:** As \( x \to -\infty \), \( e^{-0.2x} \to \infty \) and hence \[ \mathrm{f}(x) \to \infty. \] - **Sketch features:** The graph starts very high for very negative \( x \) (heading to \( +\infty \)) and decreases continuously to approach \( y = 5 \) as \( x \to \infty \). The graph is smooth and has a y-intercept at \( (0,7) \). 5. Answers based on the choices: - Increasing/Decreasing: The function is decreasing on \((-\infty, \infty)\) and never increasing. - Local extrema: There are no local extrema. Therefore, the correct choices are: For increasing/decreasing: **C. The function is decreasing on \((-\infty, \infty)\). It is never increasing.** For local extrema: **D. There are no local extrema.**

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The Deep Dive

The function \( f(x) = 5 + 2 e^{-0.2 x} \) describes a classic exponential decay model. As \( x \) increases, \( e^{-0.2 x} \) approaches zero, which means that \( f(x) \) approaches 5 from above. The graph starts at \( f(0) = 7 \) and asymptotically approaches 5 as \( x \) grows larger, resembling a horizontal shift that stabilizes. For intervals of increase and decrease, the exponential function's nature indicates that \( f(x) \) is always decreasing. Thus, you would select B: The function is increasing on (−∞, 0). It is decreasing on (0, ∞). Regarding local extrema, since \( f(x) \) has no turning points, the correct choice is D: There are no local extrema.

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