Pregunta
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Find the \( x \) and \( y \) coordinates of all infleclion points. \( I(x)=4 x^{\frac{5}{3}}+5 \) What islare the inflection poinl(s)? Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The inflection poinl(s) islare (Type an ordered pair. Use a comma to separale answers as needed.) B. There aro no inflection ponis.

Ask by Harmon Hodgson. in the United States
Mar 11,2025

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There are no inflection points.

Solución

To find the inflection points of the function \( I(x) = 4x^{\frac{5}{3}} + 5 \), we need to follow these steps: 1. **Find the second derivative of the function.** 2. **Set the second derivative equal to zero and solve for \( x \).** 3. **Determine the corresponding \( y \) values for the \( x \) values found.** 4. **Check the sign of the second derivative around the points found to confirm they are inflection points.** Let's start by finding the first and second derivatives of \( I(x) \). ### Step 1: Find the first derivative \( I'(x) \) Using the power rule, we differentiate \( I(x) \): \[ I'(x) = \frac{d}{dx}(4x^{\frac{5}{3}} + 5) = 4 \cdot \frac{5}{3} x^{\frac{5}{3} - 1} = \frac{20}{3} x^{\frac{2}{3}} \] ### Step 2: Find the second derivative \( I''(x) \) Now we differentiate \( I'(x) \): \[ I''(x) = \frac{d}{dx}\left(\frac{20}{3} x^{\frac{2}{3}}\right) = \frac{20}{3} \cdot \frac{2}{3} x^{\frac{2}{3} - 1} = \frac{40}{9} x^{-\frac{1}{3}} \] ### Step 3: Set the second derivative equal to zero To find the inflection points, we set \( I''(x) = 0 \): \[ \frac{40}{9} x^{-\frac{1}{3}} = 0 \] This equation does not have any solutions since \( x^{-\frac{1}{3}} \) cannot be zero. However, we need to check where \( I''(x) \) is undefined, which occurs when \( x = 0 \). ### Step 4: Check the sign of \( I''(x) \) We will check the sign of \( I''(x) \) around \( x = 0 \): - For \( x < 0 \): \( I''(x) = \frac{40}{9} x^{-\frac{1}{3}} \) is positive (since \( x^{-\frac{1}{3}} \) is negative). - For \( x > 0 \): \( I''(x) = \frac{40}{9} x^{-\frac{1}{3}} \) is positive (since \( x^{-\frac{1}{3}} \) is positive). Since the sign of \( I''(x) \) does not change around \( x = 0 \), there are no inflection points. ### Conclusion The correct choice is: B. There are no inflection points.

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The Deep Dive

To find the inflection points of the function \( I(x) = 4x^{\frac{5}{3}} + 5 \), we need to calculate the second derivative and determine where it changes sign. First, we find the first derivative: \[ I'(x) = \frac{d}{dx}(4x^{\frac{5}{3}}) = \frac{20}{3} x^{\frac{2}{3}} \] Next, we find the second derivative: \[ I''(x) = \frac{d}{dx}\left(\frac{20}{3} x^{\frac{2}{3}}\right) = \frac{20 \cdot \frac{2}{3}}{3} x^{-\frac{1}{3}} = \frac{40}{9} x^{-\frac{1}{3}} \] To find the inflection points, set the second derivative equal to zero: \[ \frac{40}{9} x^{-\frac{1}{3}} = 0 \] This expression is never equal to zero since \( x^{-\frac{1}{3}} \) can never be zero. However, it is undefined at \( x = 0 \). Now, we check the sign change around \( x = 0 \): - For \( x < 0 \) (e.g., \( x = -1 \)), \( I''(-1) = \frac{40}{9}(-1)^{-\frac{1}{3}} < 0 \) (concave down). - For \( x > 0 \) (e.g., \( x = 1 \)), \( I''(1) = \frac{40}{9}(1)^{-\frac{1}{3}} > 0 \) (concave up). Since there is a change from concave down to concave up at \( x = 0 \), we have an inflection point. So, the inflection point is: A. The inflection point(s) is/are (0, 5).

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