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d'y and dx? dy 8) At which of the 5 points on the graph in the figure below are dx both negative? Explain your reasoning. B

d'y and dx? dy 8) At which of the 5 points on the graph in the figure below are dx both negative? Explain your reasoning. B

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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**Question 8:** **Graph Analysis and Critical Points** At which of the 5 points on the graph in the figure below are \(\frac{dy}{dx}\) and \(\frac{d^2y}{dx^2}\) both negative? Explain your reasoning. **Graph Description:** - The graph is plotted on an XY-coordinate plane. - The graph starts at point \(O\) on the origin. - The graph proceeds to five labeled points: \(A\), \(B\), \(C\), \(D\), and \(E\). - Point \(A\) appears to be on the rising curve moving upwards. - Point \(B\) marks the peak where the graph begins to descend. - Point \(C\) is the lowest point, resembling a local minimum. - Point \(D\) is on the ascending part after \(C\). - Point \(E\) is further rising on the curve past \(D\). **Explanation:** To determine where both the first derivative (\(\frac{dy}{dx}\)) and the second derivative (\(\frac{d^2y}{dx^2}\)) are negative: 1. **\(\frac{dy}{dx}\) Negative**: This represents the slope (or rate of change) of the function. A negative slope means the function is decreasing. 2. **\(\frac{d^2y}{dx^2}\) Negative**: This indicates the concavity of the graph. When \(\frac{d^2y}{dx^2}\) is negative, the graph is concave down. Observing the graph: - **Point \(A\)**: \(\frac{dy}{dx}\) is positive (increasing slope), thus not qualifying. - **Point \(B\)**: At the peak, \(\frac{dy}{dx} = 0\). The second derivative does not determine a sign here. - **Point \(C\)**: At the trough, \(\frac{dy}{dx} = 0\). The second derivative does not determine a sign. - **Point \(D\)**: Positive slope (\(\frac{dy}{dx} > 0\)) as it is on an upward curve, so it does not qualify. - **Point \(E\)**: Positive slope \(\frac{dy}{dx} > 0\) and concave up.
Transcribed Image Text:**Question 8:** **Graph Analysis and Critical Points** At which of the 5 points on the graph in the figure below are \(\frac{dy}{dx}\) and \(\frac{d^2y}{dx^2}\) both negative? Explain your reasoning. **Graph Description:** - The graph is plotted on an XY-coordinate plane. - The graph starts at point \(O\) on the origin. - The graph proceeds to five labeled points: \(A\), \(B\), \(C\), \(D\), and \(E\). - Point \(A\) appears to be on the rising curve moving upwards. - Point \(B\) marks the peak where the graph begins to descend. - Point \(C\) is the lowest point, resembling a local minimum. - Point \(D\) is on the ascending part after \(C\). - Point \(E\) is further rising on the curve past \(D\). **Explanation:** To determine where both the first derivative (\(\frac{dy}{dx}\)) and the second derivative (\(\frac{d^2y}{dx^2}\)) are negative: 1. **\(\frac{dy}{dx}\) Negative**: This represents the slope (or rate of change) of the function. A negative slope means the function is decreasing. 2. **\(\frac{d^2y}{dx^2}\) Negative**: This indicates the concavity of the graph. When \(\frac{d^2y}{dx^2}\) is negative, the graph is concave down. Observing the graph: - **Point \(A\)**: \(\frac{dy}{dx}\) is positive (increasing slope), thus not qualifying. - **Point \(B\)**: At the peak, \(\frac{dy}{dx} = 0\). The second derivative does not determine a sign here. - **Point \(C\)**: At the trough, \(\frac{dy}{dx} = 0\). The second derivative does not determine a sign. - **Point \(D\)**: Positive slope (\(\frac{dy}{dx} > 0\)) as it is on an upward curve, so it does not qualify. - **Point \(E\)**: Positive slope \(\frac{dy}{dx} > 0\) and concave up.
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