F(x)=4+cosx;m(p/6;p)

To evaluate \( m\left(\frac{\pi}{6}; p\right) \) for the function \( f(x) = 4 + \cos x \), we first need to analyze the given function. The function \( f(x) = 4 + \cos x \) is defined for all real numbers \( x \) and involves the cosine function. The cosine function oscillates between -1 and 1, meaning: \[ \cos x \in [-1, 1] \] Hence, the range of \( f(x) \) can be determined as follows: \[ f(x) \in [4 - 1, 4 + 1] = [3, 5] \] Now, to find \( m\left(\frac{\pi}{6}; p\right) \), we assume that \( m(a; b) \) denotes the minimum value of the function over the interval \( [a, b] \). Given \( a = \frac{\pi}{6} \) and \( b = p \), we can find the minimum value of \( f(x) \) over the interval \( \left[\frac{\pi}{6}, p\right] \). 1. Calculate \( f\left(\frac{\pi}{6}\right) \): \[ f\left(\frac{\pi}{6}\right) = 4 + \cos\left(\frac{\pi}{6}\right) = 4 + \frac{\sqrt{3}}{2} \] 2. Evaluate \( f(x) \) at the endpoints and look for critical points where the derivative could be 0. 3. The derivative \( f'(x) \) is: \[ f'(x) = -\sin x \] Set \( f'(x) = 0 \): \(-\sin x = 0\) gives \( x = n\pi \) for integers \( n \). However, only consider values within our interval \( \left[\frac{\pi}{6}, p\right] \). 4. Depending on the value of \( p \), we will also need to evaluate \( f(p) \). Finally, take the minimum of \( f\left(\frac{\pi}{6}\right) \) and \( f(p) \). **Summary:** - Compute \( f\left(\frac{\pi}{6}\right) \). - Evaluate \( f(p) \). - The minimum value \( m\left(\frac{\pi}{6}; p\right) = \min\left(f\left(\frac{\pi}{6}\right), f(p)\right) \). If you have a specific value for \( p \) or want further assistance or clarification, please provide it!