Final answer:
To find the x-values at which f'(x) is zero or undefined for f(x) = 3x√x - 9, we need to take the derivative of f(x) and set it equal to zero. The derivative is found to be f'(x) = 6√x, and it is zero when x = 0. The derivative is always defined for all x-values greater than or equal to zero.
Step-by-step explanation:
To find the x-values at which f'(x) is zero or undefined for f(x) = 3x√x - 9, we need to find the derivative of f(x) which is f'(x). Let's start by finding f'(x) using the power rule and the product rule:
f'(x) = d/dx(3x√x - 9) = (3√x + 3x(√x)' - 0) = 3√x + 3x(1/2)x^(-1/2) = 3√x + 3x√x/x = 3√x + 3√x = 6√x
To find the x-values at which f'(x) is zero, we set 6√x = 0 and solve for x: 6√x = 0 => √x = 0 => x = 0
Since the square root of zero is defined, f'(x) is zero at x = 0. To find the x-values at which f'(x) is undefined, we need to find where the derivative is not defined. In this case, the derivative is always defined for all x-values greater than or equal to zero. Therefore, f'(x) is defined for all x-values, and there are no x-values at which it is undefined.