Problem 46 - Thomas Calculus

Question

In Exercises \(41-50,\) determine all critical points for each function. $$ f(x)=\frac{x^{2}}{x-2} $$

Step-by-Step Solution

Answer

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

Step 1: Find the derivative

To find the critical points, we first need to determine the derivative of the function. The given function is \(f(x) = \frac{x^2}{x-2}\). We apply the quotient rule, which states that if \(f(x) = \frac{u(x)}{v(x)}\), then \(f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}\). Here, \(u(x) = x^2\) and \(v(x) = x-2\). Calculate \(u'(x) = 2x\) and \(v'(x) = 1\). Hence, \[f'(x) = \frac{(2x)(x-2) - (x^2)(1)}{(x-2)^2} = \frac{2x^2 - 4x - x^2}{(x-2)^2} = \frac{x^2 - 4x}{(x-2)^2}.\]

Step 2: Set the derivative to zero

To find critical points, set the derivative equal to zero and solve for \(x\). This involves solving the equation \[\frac{x^2 - 4x}{(x-2)^2} = 0.\] Since a fraction is zero when the numerator is zero (and the denominator is non-zero), solve \[x^2 - 4x = 0.\] Factor the numerator: \[x(x - 4) = 0.\] So, \(x = 0\) or \(x = 4\) are solutions, provided they don't make the denominator zero.

Step 3: Check the domain of the function

The function \(f(x) = \frac{x^2}{x-2}\) is undefined where the denominator is zero, i.e., where \(x - 2 = 0\), so \(x = 2\). Therefore, \(x = 2\) is not in the domain of the function and cannot be a critical point.

Step 4: Identify the critical points

The critical points are the solutions from Step 2 that lie within the domain of the function. From Step 2, we derived \(x = 0\) and \(x = 4\). Since both are within the domain, these are the critical points of the function.

Key Concepts

Derivative Calculus

In the realm of calculus, the concept of derivatives is integral to understanding how functions behave. Derivatives tell us the rate at which a function is changing at any given point. They are the mathematical expressions indicating the steepness or slope of a function's graph at a point. For the given function, which is a fraction, the derivative is essential in establishing the function's critical points, where the slope is zero or undefined.

In this exercise, we need to derive the function \(f(x) = \frac{x^2}{x-2}\). The derivative helps identify points where the rate of change of the function equals zero, which are potential critical points. By applying the quotient rule, which is necessary for taking derivatives of fractions, we can calculate this derivative.

Quotient Rule

The quotient rule in calculus is specifically used for differentiating functions that are ratios of two other functions. So if you encounter a function that looks like \(\frac{u(x)}{v(x)}\), you can recall that the derivative \(f'(x)\) is given by \(\frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}\). This rule is crucial when handling problems with fractional functions, as seen in this exercise.

For instance, in our problem, we identified the numerator function as \(u(x) = x^2\) and the denominator as \(v(x) = x-2\). Using the quotient rule, we can compute \(u'(x) = 2x\) and \(v'(x) = 1\), letting us find the derivative: \(f'(x) = \frac{2x^2 - 4x - x^2}{(x-2)^2} = \frac{x^2 - 4x}{(x-2)^2}\). This result is crucial for finding the critical points.

Function Analysis

Analyzing a function involves identifying critical points, which can be found through the function's derivative. A critical point occurs where the derivative equals zero or is undefined. For the function \(f(x) = \frac{x^2}{x-2}\), we derived \(f'(x) = \frac{x^2 - 4x}{(x-2)^2}\).

Next, we set the derivative to zero: \(\frac{x^2 - 4x}{(x-2)^2} = 0\). A fraction equals zero when its numerator is zero, provided the denominator isn't zero. Thus, solving \(x^2 - 4x = 0\), we find \(x = 0\) and \(x = 4\). Both values do not make the denominator \((x-2)^2\) zero, showing these are potential critical points. Checking the domain, \(x = 2\) is excluded, keeping us from incorrectly identifying it as a critical point.

Solving Equations

Solving equations is a vital skill in calculus, particularly in finding critical points by setting derivatives to zero. In our problem, the equation \(x^2 - 4x = 0\) was obtained by isolating the numerator of the derivative. Solving such equations is key to identifying critical points of a function.

We factored the equation into \(x(x-4) = 0\), leading to the solutions \(x = 0\) and \(x = 4\). These solutions represent the actual critical points of the function, as they lie within the domain not invalidated by the denominator. Understanding solving techniques, such as factoring, ensures you can find these points efficiently and accurately.