Does the graph of the furiction below have any horizontal tangents in the interval \( 0 \leq x \leq 2 \pi \) ? If so, where? If not, why not? Visualize your findings by graphing the function with a grapher. \( y=4 x+8 \sin x \) - A. The function has a horizontal tangent at \( x=\square \). (Type your answer in radians. Type an exact answer in terms of \( \pi \). Use a comma to separate answers as needed.) B. The function has no horizontal tangents in the interval \( 0 \leq x \leq 2 \pi \) because it is never zero in the interval \( 0 \leq x \leq 2 \pi \). C. The function has no horizontal tangents in the interval \( 0 \leq x \leq 2 \pi \) because its derivative is never zero in the interval \( 0 \leq x \leq 2 \pi \). D. The function has no horizontal tangents in the interval \( 0 \leq x \leq 2 \pi \) because it is defined at every point in the interval \( 0 \leq x \leq 2 \pi \).

To determine if the function \( y = 4x + 8 \sin x \) has any horizontal tangents in the interval \( 0 \leq x \leq 2\pi \), we need to find the derivative of the function and check where it is equal to zero. 1. **Find the derivative**: \[ y' = \frac{d}{dx}(4x + 8 \sin x) = 4 + 8 \cos x \] 2. **Set the derivative equal to zero** to find horizontal tangents: \[ 4 + 8 \cos x = 0 \] \[ 8 \cos x = -4 \] \[ \cos x = -\frac{1}{2} \] 3. **Find the values of \( x \)** where \( \cos x = -\frac{1}{2} \) in the interval \( 0 \leq x \leq 2\pi \): The cosine function is equal to \(-\frac{1}{2}\) at: \[ x = \frac{2\pi}{3}, \quad \text{and} \quad x = \frac{4\pi}{3} \] Thus, the function has horizontal tangents at: \[ x = \frac{2\pi}{3}, \frac{4\pi}{3} \] ### Answers: A. The function has a horizontal tangent at \( x = \frac{2\pi}{3}, \frac{4\pi}{3} \). B. The function has no horizontal tangents in the interval \( 0 \leq x \leq 2\pi \) because it is never zero in the interval \( 0 \leq x \leq 2\pi \). (This statement is incorrect.) C. The function has no horizontal tangents in the interval \( 0 \leq x \leq 2\pi \) because its derivative is never zero in the interval \( 0 \leq x \leq 2\pi \). (This statement is incorrect.) D. The function has no horizontal tangents in the interval \( 0 \leq x \leq 2\pi \) because it is defined at every point in the interval \( 0 \leq x \leq 2\pi \). (This statement is incorrect.) ### Conclusion: The correct answer is that the function has horizontal tangents at \( x = \frac{2\pi}{3} \) and \( x = \frac{4\pi}{3} \).