Problem 44 - Thomas Calculus

Question

In Exercises \(41-50,\) determine all critical points for each function. $$ g(x)=(x-1)^{2}(x-3)^{2} $$

Step-by-Step Solution

Answer

Critical points for \( g(x) \) are \( x = 1, 2, \) and \( 3 \).

Step 1: Find the Derivative

To find the critical points of the function \( g(x) = (x-1)^{2}(x-3)^{2} \), we first need to find the derivative \( g'(x) \). Use the product rule where if \( u = (x-1)^2 \) and \( v = (x-3)^2 \), then \( g'(x) = u'v + uv' \). Begin by differentiating \( u \) and \( v \). The derivative of \( u = (x-1)^2 \) is \( u' = 2(x-1) \), and the derivative of \( v = (x-3)^2 \) is \( v' = 2(x-3) \). Substitute back into the product rule.

Step 2: Substitute into the Product Rule

Using the derivatives found in Step 1, substitute into the product rule formula: \[ g'(x) = 2(x-1)(x-3)^2 + (x-1)^2 \, 2(x-3) \]Simplify the expression to find \( g'(x) \).

Step 3: Simplify the Derivative

Expand and combine like terms in the expression found in Step 2:\[ g'(x) = 2(x-1)(x-3)^2 + 2(x-1)^2(x-3) \]Distribute and combine terms:\[ g'(x) = 2(x - 1)(x^2 - 6x + 9) + 2(x^2 - 2x + 1)(x - 3) \]Continue expanding and combining like terms.

Step 4: Set Derivative Equal to Zero

After simplifying, set the derivative \( g'(x) \) to zero to find the critical points:\[ g'(x) = (x-1)(x-3)(2x - 4) = 0 \]The critical points occur where this product is equal to zero, solving the equation gives the potential critical points.

Step 5: Solve for Critical Points

The equation \( (x-1)(x-3)(2x - 4) = 0 \) implicates three factors that each can be zero:1. \( x-1=0 \) gives \( x = 1 \).2. \( x-3=0 \) gives \( x = 3 \).3. \( 2x - 4=0 \) gives \( x = 2 \).Thus, the critical points are \( x = 1, 2, \) and \( 3 \).

Step 6: Verify with Original Function

Check the function at the identified critical points to verify if the behavior changes — whether it’s a local maximum, minimum, or a point of inflection.

Key Concepts

Derivative

In calculus, the derivative of a function measures how the function's output changes as its input changes. It's like finding the slope of a curve at any given point. Derivatives are fundamental to understanding a function's rate of change and are expressed using the symbol \( f'(x) \) or \( \frac{df}{dx} \).
To find the derivative of a polynomial function, like the one in our exercise, we often rely on rules such as the power rule, chain rule, and product rule. For our specific function, \( g(x) = (x-1)^2(x-3)^2 \), the product rule is crucial because the function is a product of two separate expressions.
The goal of finding the derivative here is to establish where the function's rate of change becomes zero, indicating potential critical points where the behavior might shift dramatically, possibly achieving local maximums or minimums.

Product Rule

The product rule is specifically used to differentiate functions that are the product of two or more terms. It allows us to break down the process into simpler steps. If you have two functions, \( u(x) \) and \( v(x) \), their product's derivative, \( (uv)' \), is calculated as \( u'v + uv' \).
In our exercise, the function \( g(x) = (x-1)^2(x-3)^2 \) is such a case. We let \( u = (x-1)^2 \) and \( v = (x-3)^2 \). We found \( u' = 2(x-1) \) and \( v' = 2(x-3) \) through simple power rule application. Using the product rule, we found that the derivative is:

• \( g'(x) = 2(x-1)(x-3)^2 + (x-1)^2 \cdot 2(x-3) \)

This step is critical as it simplifies finding where \( g(x) \) is stationary, often leading us to the critical numbers.

Polynomial Function

Polynomial functions are expressions consisting of variables raised to whole number powers and multiplied by constants. They are some of the simplest types of functions and are denoted by terms like \( ax^n + bx^{n-1} + ... + a_0 \). In our case, \( g(x) = (x-1)^2 (x-3)^2 \) simplifies into a polynomial form when expanded.
Polynomials are essential in calculus because they are smooth and continuous, making them ideal candidates for differentiation and integration. The degree of a polynomial gives insights into the number of roots and critical points it could have. Each factor, such as \((x-1)\) or \((x-3)\), indicates potential critical points when using calculus techniques to simplify and solve such equations.
In solving our exercise, the polynomial representation helped us apply the derivative and product rule effectively.

Critical Numbers

Critical numbers of a function are the values of \( x \) at which the derivative is zero or undefined. These points represent the potential locations where a function’s rate of change is zero, indicating possible local maxima, minima, or inflection points.
For our function \( g(x) = (x-1)^2(x-3)^2 \), calculating the derivative and setting it equal to zero allowed us to solve for the critical numbers. After simplifying the derivative expression to \( (x-1)(x-3)(2x-4) \), we found the critical points by solving:

• \( x-1 = 0 \rightarrow x = 1 \)
• \( x-3 = 0 \rightarrow x = 3 \)
• \( 2x-4 = 0 \rightarrow x = 2 \)

These critical numbers help us understand where the function might attain its highest or lowest values in certain intervals.