Consider the function $f(x)=6(x-4)^{2 / 3}$. For this function there are two important intervals: $(-\infty, A)$ and $(A, \infty)$ where $A$ is a critical number.
$A$ is $\square$
Answer (1)
Find the derivative of the function: f ′ ( x ) = ( x − 4 ) 3 1 4 .
Set the derivative equal to zero and solve for x . There are no solutions.
Find where the derivative is undefined by setting the denominator equal to zero: ( x − 4 ) 3 1 = 0 , which gives x = 4 .
The critical number is therefore 4 .
Explanation
Problem Analysis We are given the function f ( x ) = 6 ( x − 4 ) 2/3 and asked to find the critical number A such that the intervals ( − ∞ , A ) and ( A , ∞ ) are important intervals for the function. Critical numbers occur where the derivative of the function is either equal to zero or undefined. So, we need to find the derivative of f ( x ) and determine where it is zero or undefined.
Finding the Derivative First, let's find the derivative of f ( x ) with respect to x . Using the power rule and chain rule, we have: f ′ ( x ) = 6 ⋅ 3 2 ( x − 4 ) 3 2 − 1 = 4 ( x − 4 ) − 3 1 = ( x − 4 ) 3 1 4
Finding Where Derivative is Zero Now, we need to find where the derivative is equal to zero or undefined. Setting f ′ ( x ) = 0 , we have: ( x − 4 ) 3 1 4 = 0 This equation has no solution, as a fraction can only be zero if its numerator is zero, but the numerator here is 4.
Finding Where Derivative is Undefined Next, we need to find where the derivative is undefined. The derivative is undefined when the denominator is equal to zero: ( x − 4 ) 3 1 = 0 Raising both sides to the power of 3, we get: x − 4 = 0 Solving for x , we find: x = 4 Thus, the derivative is undefined at x = 4 .
Final Answer Therefore, the critical number A is 4. The important intervals are ( − ∞ , 4 ) and ( 4 , ∞ ) .
Conclusion The critical number is A = 4 .
Examples
In optimization problems, finding critical points is essential. For example, if you're designing a container and want to minimize the surface area for a given volume, you'd find the critical points of the surface area function to determine the dimensions that achieve this minimum. Similarly, in economics, critical points can help determine the production level that maximizes profit or minimizes cost.
