Problem 3 - Calculus Early Transcendentals

Question

What are the conditions for a critical point of a function \(f ?\)

Step-by-Step Solution

Answer

Answer: The conditions for a critical point of a function are: 1. The derivative f'(x) = 0 (stationary point). 2. The derivative f'(x) is undefined (singular point).

Step 1: Definition of a critical point

A critical point of a function \(f\) is a point on its domain where the derivative of the function is either equal to zero or is undefined. There are two types of critical points: stationary points and singular points.

Step 2: Identifying stationary points

To identify stationary points, we find the first derivative of the function \(f'(x)\), then solve for x when \(f'(x) = 0\). If a function can have a local maximum or minimum, it occurs at a stationary point. For example, if \(f(x) = x^3 - 6x^2 + 12x\), find its derivative: $$ f'(x) = \frac{d}{dx}(x^3 - 6x^2 + 12x) = 3x^2 - 12x + 12 $$ Now, set the first derivative equal to zero and solve for x: $$ 3x^2 - 12x + 12 = 0 $$

Step 3: Identifying singular points

To identify singular points, we look for points on the domain of the function where the derivative \(f'(x)\) is not defined. These are usually points where the function is not continuous or not differentiable. For example, if \(f(x) = \sqrt{x}\), find its derivative: $$ f'(x) = \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}} $$ In this case, \(f'(x)\) is not defined at \(x=0\), since dividing by zero is undefined. This means that \(x=0\) is a singular point. In conclusion, the conditions for a critical point of a function \(f\) are: 1. The derivative \(f'(x) = 0\) (stationary point) 2. The derivative \(f'(x)\) is undefined (singular point)