Determine whether the Mean value theorem can be appiled to \( f \) on the closed interval \( [a, b] \). (Select ali that apply.) \[ f(x)=\frac{x+4}{x-2},[-6,6] \] Yes, the Mean Value theorem can be applied. No, \( f \) is not continuous on \( [a, b] \). No, \( f \) is not differentiable on ( \( a, b \) ). None of the above. If the Mean Value theorem can be applied, find all values of \( c \) in the open interval \( (a, b) \) such that \( f^{\prime}(c)=\frac{f(b)-f(a)}{b-a} \). (Enter your answers as a comma-separated list. If the Mean value theorem cannot be applied, enter NA.) \( c=\square \)

To determine if the Mean Value Theorem (MVT) can be applied to the function \( f(x) = \frac{x+4}{x-2} \) on the interval \( [-6, 6] \), we first check continuity and differentiability. The function has a vertical asymptote at \( x = 2 \), meaning it is not continuous on the closed interval \( [-6, 6] \). Therefore, the MVT cannot be applied. For the value of \( c \), since MVT does not apply: \( c = NA \). Now, let's delve into the fascinating world of calculus! 😄 The Mean Value Theorem is not just a theoretical concept; it’s a bridge connecting instantaneous rates of change (derivatives) to average rates of change over intervals. This connection has profound implications in understanding motion, optimization, and curve sketching. Also, have you ever thought about why the MVT requires both continuity and differentiability? It’s simple yet crucial! Continuity ensures that there’s no "jump" in the function, while differentiability means the function isn’t too "jumpy" at infinitesimally small scales. This harmony is what allows us to find that sweet spot where instantaneous and average rates meet! 🌀