Consider the function f(x)=4(x-5)^(2/3). For this function, there are two important intervals: (-Inf, A) and (A, Inf) Where A is a critical number. Find A
we know that at x=5 the tangent is vertical, so A=5
To find the critical number A for the function f(x) = 4(x - 5)^(2/3), we need to find the value of x that makes the derivative of the function equal to zero. The derivative of f(x) can be found using the power rule for differentiation and the chain rule.
1. Start by finding the derivative of the function f(x):
f'(x) = d/dx [4(x - 5)^(2/3)]
2. Apply the power rule for differentiation:
f'(x) = 4 * (2/3) * (x - 5)^(2/3 - 1) * (d/dx)(x - 5)
3. Simplify the expression:
f'(x) = 8/3 * (x - 5)^(-1/3) * 1
4. Now, set f'(x) equal to zero and solve for x:
0 = 8/3 * (x - 5)^(-1/3)
To make the equation simpler, we can multiply both sides of the equation by 3/8:
0 * (3/8) = 8/3 * (x - 5)^(-1/3) * (3/8)
0 = (2/3) * (x - 5)^(-1/3)
Since we cannot divide by zero, for this equation to hold, the expression (x - 5)^(-1/3) should be undefined, which occurs when the base is zero.
(x - 5)^(-1/3) is undefined when x - 5 = 0.
So, x - 5 = 0 or x = 5.
Therefore, the critical number A for the function f(x) = 4(x - 5)^(2/3) is A = 5.