Question

In Exercises 9-16, find any critical numbers of the function. $$ f(x)=x^{2}(x-3) $$

Short answer

The critical points are \(x = 0\) and \(x = 2\).

Step-by-step solution

Find the derivative

The function is \( f(x) = x^2(x-3) \). Applying the product rule for differentiation, which states that the derivative of a product of two functions is the derivative of the first times the second, plus the first times the derivative of the second, we get: \( f'(x) = 2x(x-3) + x^2(1) = 2x^2 - 6x + x^2 = 3x^2 - 6x \).

Solve for derivative equals to zero

Set the derivative equal to zero and solve for \(x\): \( 3x^2 - 6x = 0 \). Factoring \x\ out, we get: \( x(3x - 6) = 0 \).

Identify the critical numbers

Now, set the result from Step 2 to zero: \(x = 0\) and \(3x-6 = 0\). If we solve for \(x\), we get \(x = 2\). The critical numbers are the roots of the derivative, which are \(x = 0\) and \(x = 2\).