Consider the function $f(x)=5(x-4)^{2 / 3}$. For this function there are two important intervals: $(-\infty, A)$ and $(A, \infty)$ where $A$ is a critical number. $A$ is
Solution
Step 1 ：Consider the function \(f(x)=5(x-4)^{2 / 3}\). For this function there are two important intervals: \((-\infty, A)\) and \((A, \infty)\) where \(A\) is a critical number.
Step 2 ：The critical numbers of a function are the x-values where the derivative of the function is either zero or undefined. To find the critical number for this function, we need to first find its derivative.
Step 3 ：The derivative of the function \(f(x)=5(x-4)^{2 / 3}\) is \(f'(x) = \frac{10}{3}(x - 4)^{-1/3}\).
Step 4 ：The derivative of the function is undefined at \(x = 4\). This is because the denominator of the derivative becomes zero at this point, leading to a division by zero error.
Step 5 ：Therefore, the critical number \(A\) is 4.
Step 6 ：Final Answer: \(A = \boxed{4}\)
