Questions: a. Find the local extrema of the function f(x) = sin 3x on the interval 0 ≤ x ≤ 2π/3, and say where they occur. b. Graph the function and its derivative together. Comment on the behavior of f in relation to the signs and values of f′.

Solution Steps

The given function is \( f(x) = \sin 3x \). To find the local extrema, we first need to find its derivative: \[ f'(x) = \frac{d}{dx} (\sin 3x) = 3 \cos 3x \]

To find the critical points, we set the derivative equal to zero: \[ 3 \cos 3x = 0 \] \[ \cos 3x = 0 \]

The cosine function is zero at \( \frac{\pi}{2} + k\pi \) for integer \( k \). Therefore: \[ 3x = \frac{\pi}{2} + k\pi \] \[ x = \frac{\pi}{6} + \frac{k\pi}{3} \]

We need to find the values of \( x \) within the interval \( 0 \leq x \leq \frac{2\pi}{3} \): For \( k = 0 \): \[ x = \frac{\pi}{6} \] For \( k = 1 \): \[ x = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2} \] For \( k = 2 \): \[ x = \frac{\pi}{6} + \frac{2\pi}{3} = \frac{5\pi}{6} \] (This is outside the interval)

Evaluate \( f(x) \) at \( x = 0 \), \( x = \frac{\pi}{6} \), \( x = \frac{\pi}{2} \), and \( x = \frac{2\pi}{3} \): \[ f(0) = \sin 0 = 0 \] \[ f\left(\frac{\pi}{6}\right) = \sin\left(\frac{\pi}{2}\right) = 1 \] \[ f\left(\frac{\pi}{2}\right) = \sin\left(\frac{3\pi}{2}\right) = -1 \] \[ f\left(\frac{2\pi}{3}\right) = \sin(2\pi) = 0 \]

Final Answer