Pregunta
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Find the intervals on which the graph of \( f \) is concave upward, the intervals on which the graph of \( f \) is concave downward, and the inflection points. \( f(x)=\ln \left(x^{2}-2 x+10\right) \) For what interval(s) of \( x \) is the graph of \( f \) concave upward? Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. B. The graph is never concave upward. For what interval(s) of \( x \) is the graph of \( f \) concave downward? Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. \( \square \) (Type your answer in interval notation. Type an exact answer. Use a comma to separate answers as needed.) B. The graph is never concave downward. Determine the \( x \) coordinates of any inflection points of the graph of \( f(x) \). Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. \( x=\square \) (Type an exact answer. Use a comma to separate answers as needed.) A comma to separate answers as needed.) A

Ask by Rowe Garza. in the United States
Mar 11,2025

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The graph of \( f(x) = \ln(x^2 - 2x + 10) \) is concave upward on the interval \( (-2, 4) \) and has inflection points at \( x = -2 \) and \( x = 4 \).

Solución

To find the intervals on which the graph of \( f(x) = \ln(x^2 - 2x + 10) \) is concave upward, concave downward, and the inflection points, we need to find the second derivative of the function and analyze its sign. Given function: \( f(x) = \ln(x^2 - 2x + 10) \) First, let's find the first derivative of the function: \[ f'(x) = \frac{2x - 2}{x^2 - 2x + 10} \] Next, let's find the second derivative of the function: \[ f''(x) = \frac{(x^2 - 2x + 10)(2) - (2x - 2)(2x - 2)}{(x^2 - 2x + 10)^2} \] Simplifying the second derivative: \[ f''(x) = \frac{2x^2 - 4x + 20 - 4x^2 + 8x - 4}{(x^2 - 2x + 10)^2} \] \[ f''(x) = \frac{-2x^2 + 4x + 16}{(x^2 - 2x + 10)^2} \] Now, we need to find the intervals on which the graph of \( f \) is concave upward and concave downward by analyzing the sign of the second derivative. Concave upward: \( f''(x) > 0 \) Concave downward: \( f''(x) < 0 \) Let's analyze the sign of the second derivative to determine the intervals on which the graph of \( f \) is concave upward and concave downward. Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: \(-2x^{2}+4x+16>0\) - step1: Rewrite the expression: \(-2x^{2}+4x+16=0\) - step2: Factor the expression: \(-2\left(x-4\right)\left(x+2\right)=0\) - step3: Divide the terms: \(\left(x-4\right)\left(x+2\right)=0\) - step4: Separate into possible cases: \(\begin{align}&x-4=0\\&x+2=0\end{align}\) - step5: Solve the equation: \(\begin{align}&x=4\\&x=-2\end{align}\) - step6: Determine the test intervals: \(\begin{align}&x<-2\\&-24\end{align}\) - step7: Choose a value: \(\begin{align}&x_{1}=-3\\&x_{2}=1\\&x_{3}=5\end{align}\) - step8: Test the chosen value: \(\begin{align}&x<-2\textrm{ }\textrm{is not a solution}\\&-24\textrm{ }\textrm{is not a solution}\end{align}\) - step9: The final solution is \(-2

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To analyze the concavity of the function \( f(x) = \ln(x^2 - 2x + 10) \), we start by finding its second derivative. 1. **First Derivative**: \[ f'(x) = \frac{1}{x^2 - 2x + 10}(2x - 2) \] 2. **Second Derivative**: To find \( f''(x) \), we apply the quotient and product rules which become a bit tedious, but ultimately, after some simplification, you will find the sign of \( f''(x) \) gives us insight into the concavity of the function. Upon finding \( f''(x) \), you will determine the intervals of concavity by checking the intervals between critical points where \( f''(x) = 0 \) or is undefined. ### For the Concavity: 1. **Concave Upward**: After reviewing the sign of the second derivative, you will find that \( f(x) \) is concave upward on the interval \( (-\infty, 1) \cup (1, \infty) \). 2. **Concave Downward**: The function is concave downward where \( f''(x) < 0 \). Upon evaluating the conditions derived from the second derivative, you will find that there are no intervals where the function is concave downward. ### For Inflection Points: An inflection point occurs where the concavity changes. By setting \( f''(x) = 0 \) and solving, you may find an \( x \)-coordinate of an inflection point at \( x = 1 \). Summarizing the results: - For concave upward: The graph is concave upward on the intervals \( (-\infty, 1) \cup (1, \infty) \). - For concave downward: The graph is never concave downward. - Inflection points: \( x = 1 \) is the inflection point.

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