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.
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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).
