Problem 50 - University Calculus: Early Transcendentals

Question

Determine all critical points for each function. $$f(x)=\frac{x^{2}}{x-2}$$

Step-by-Step Solution

Answer

The critical points are at \( x = 0 \), \( x = 4 \), and \( x = 2 \).

Step 1 - Define Critical Points

Critical points of a function are points where the derivative is zero or undefined. For a function \( f(x) \), these occur where \( f'(x) = 0 \) or \( f'(x) \) does not exist.

Step 2 - Find the Derivative

The function given is \( f(x) = \frac{x^2}{x-2} \). To find its derivative, use the quotient rule for derivatives: \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \), where \( u = x^2 \) and \( v = x-2 \). So, \( u' = 2x \) and \( v' = 1 \). Apply the formula:\[ f'(x) = \frac{2x(x-2) - x^2(1)}{(x-2)^2} = \frac{2x^2 - 4x - x^2}{(x-2)^2} = \frac{x^2 - 4x}{(x-2)^2} \].

Step 3 - Set Derivative to Zero

Set the derivative equal to zero to find critical points where the derivative is zero:\[ \frac{x^2 - 4x}{(x-2)^2} = 0 \].The numerator must be zero, thus \( x^2 - 4x = 0 \). Factor the expression: \( x(x - 4) = 0 \). So, \( x = 0 \) or \( x = 4 \) are potential critical points.

Step 4 - Check Where Derivative is Undefined

The derivative \( f'(x) = \frac{x^2 - 4x}{(x-2)^2} \) is undefined where the denominator \((x-2)^2\) is zero, which is at \( x = 2 \). Thus, \( x = 2 \) is another critical point due to the function's derivative being undefined here.

Step 5 - Summary of Critical Points

The critical points of the function are the values of \( x \) where \( f'(x) = 0 \) or where the derivative is undefined. Thus, the critical points are \( x = 0 \), \( x = 4 \), and \( x = 2 \).

Key Concepts

Derivatives

Derivatives help us understand how a function changes or how it behaves at a particular point. They provide the rate of change or the slope of a function at any given point.
When we work with derivatives, we are essentially asking, "How does the function move as we change the input?" In our example, the function is given as \( f(x) = \frac{x^2}{x-2} \).
To find its derivative, we turn to calculus rules such as the quotient rule (which we will revisit shortly). This derivative will indicate where the slope of our function becomes zero or undefined -- these are our critical points.
Knowing derivatives is more than just finding slopes; it involves examining the behavior of a curve, finding maximum or minimum points, and understanding the function's unknown nature at specific stages.

Quotient Rule

The quotient rule is a handy tool in calculus when dealing with quotients or divisions of two functions, such as \( \frac{u}{v} \). In our exercise, \( u = x^2 \) and \( v = x-2 \).
To differentiate this division, we use the formula: \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \). It might seem a bit complex at first, but let's break it down:

• Find \( u' \) which is the derivative of the numerator. Here, that's \( u' = 2x \).
• Find \( v' \), the derivative of the denominator. For us, \( v' = 1 \).
• Substitute these into the quotient rule formula to get \( f'(x) = \frac{2x(x-2) - x^2(1)}{(x-2)^2} \).
• Simplify the expression to \( \frac{x^2 - 4x}{(x-2)^2} \).

Mastering the quotient rule allows you to handle any function that can be expressed as a ratio, a common situation in calculus. It provides a way to tackle complex functions with functional layers of the numerator and denominator.

Undefined Functions

There are instances in mathematics where a function or its derivative is undefined. This typically happens where the function's denominator equals zero (resulting in division by zero).
For our function's derivative, \( f'(x) = \frac{x^2 - 4x}{(x-2)^2} \), the derivative becomes undefined at any point where the denominator \((x-2)^2\) is zero. Thus, the derivative is undefined at \( x = 2 \).
This is crucial in identifying critical points since a derivative being undefined indicates a potential for a steep slope or vertical tangent, possibly marking a critical change in behavior of the graph.
It's important to find these points since they often denote boundaries or asymptotes in the graph of a function, contributing to our full understanding of its character and how it behaves across its domain.