Questions: Find the absolute extrema of the function on the closed interval. y=9 cos(x), [0,2 π]

Solution Steps

To find the absolute extrema of the function \( y = 9 \cos(x) \) on the interval \([0, 2\pi]\), we need to evaluate the function at critical points and endpoints. First, find the derivative of the function and set it to zero to find critical points. Then, evaluate the function at these critical points and at the endpoints \( x = 0 \) and \( x = 2\pi \). Compare these values to determine the absolute minimum and maximum.

To find the critical points of the function \( y = 9 \cos(x) \), we first compute its derivative: \[ \frac{dy}{dx} = -9 \sin(x) \] Setting the derivative equal to zero gives: \[ -9 \sin(x) = 0 \implies \sin(x) = 0 \] The solutions in the interval \([0, 2\pi]\) are \( x = 0 \) and \( x = 2\pi \).

Next, we evaluate the function at the critical points and the endpoints of the interval: \[ f(0) = 9 \cos(0) = 9 \] \[ f(2\pi) = 9 \cos(2\pi) = 9 \] Since the only critical point found is \( x = 0 \), we have: \[ f(0) = 9 \quad \text{and} \quad f(2\pi) = 9 \]

The values of the function at all evaluated points are: \[ f(0) = 9, \quad f(2\pi) = 9 \] Thus, both the minimum and maximum values of the function on the interval \([0, 2\pi]\) are: \[ \text{Minimum: } (x, y) = (0, 9) \quad \text{and} \quad (x, y) = (2\pi, 9) \] \[ \text{Maximum: } (x, y) = (0, 9) \quad \text{and} \quad (x, y) = (2\pi, 9) \]

Final Answer