Problem 24 - Calculus of a Single Variable

Question

Use the derivative to determine whether the function is strictly monotonic on its entire domain and therefore has an inverse function. \(f(x)=x^{3}-6 x^{2}+12 x\)

Step-by-Step Solution

Answer

The function \(f(x) = x^{3} - 6x^{2} + 12x\) is not strictly monotonic over its whole domain as there are points where its derivative changes sign, hence it does not have an inverse function.

Step 1: Find the derivative of the function

The formula for the derivative of any function \(f(x)\) is \(f'(x)\).\nThe derivative of the function \(f(x) = x^{3} - 6x^{2} + 12x\) is found using the power rule. Apply the rule, which states that the derivative of \(x^{n}\) is \(n*x^{n-1}\):\nDerivative \(f'(x) = 3x^{2} - 12x + 12\).

Step 2: Analyze the derivative

To have a strictly monotonic function, the function's derivative must always be positive or always be negative over its entire domain.\nTo check that, let's first find the critical points of \(f'(x)\) by setting \(f'(x)= 0\) and solving the equation for x:\nSetting \(f'(x) = 0\) gives \(3x^{2} - 12x + 12 = 0\) . Solving that quadratic equation gives \(x = 2\pm \sqrt{2}\), which are the critical points of \(f'(x)\).\nThen, substitute any number greater than the larger critical point, any number between the critical points and any number lesser than the smaller critical point into \(f'(x)\) respectively, and analyze the sign.\nPick \(x = 3\) which is greater than \(2 + \sqrt{2}\), then \(f'(3) > 0 \).\nPick \(x = 2\) which is between \(2 - \sqrt{2}\) and \(2 + \sqrt{2}\), then \(f'(2) < 0 \).\nPick \(x = 1\) which is lesser than \(2 - \sqrt{2}\), then \(f'(1) > 0 \).\nThe function is not strictly monotonic because there are sub-intervals where the derivative switches sign (from positive to negative or vice versa). Therefore, it does not have an inverse function.