Problem 44 - Thomas Calculus in SI Units

Question

Determine all critical points for each function. $$g(x)=(x-1)^{2}(x-3)^{2}$$

Step-by-Step Solution

Answer

The critical points are at \( x = 1, 2, \text{ and } 3 \).

Step 1: Find the First Derivative

To find the critical points, we must first determine where the derivative of the function is zero or undefined. We begin by finding the derivative of the function \( g(x) = (x-1)^2(x-3)^2 \). Note that this function is a product of two expressions. Use the product rule: if \( u(x) = (x-1)^2 \) and \( v(x) = (x-3)^2 \), then the derivative \( g'(x) = u'(x)v(x) + u(x)v'(x) \).

Step 2: Apply the Product Rule

First, find \( u'(x) \) and \( v'(x) \):\[ u(x) = (x-1)^2 \Rightarrow u'(x) = 2(x-1) \]\[ v(x) = (x-3)^2 \Rightarrow v'(x) = 2(x-3) \]Using the product rule: \[ g'(x) = [2(x-1)(x-3)^2] + [(x-1)^2 \, 2(x-3)] \] Simplify this expression.

Step 3: Simplify the Derivative

Simplify \( g'(x) = 2(x-1)(x-3)^2 + 2(x-1)^2(x-3) \):Combine like terms:\[ 2(x-1)(x-3)(x-3 + x-1) = 2(x-1)(x-3)(2x-4) \]Factor further if possible.

Step 4: Solve \( g'(x) = 0 \)

Set the derivative equal to zero to find the critical points:\[ 2(x-1)(x-3)(2x-4) = 0 \]This equation equals zero when any of the factors equals zero:1. \( x-1 = 0 \Rightarrow x = 1 \)2. \( x-3 = 0 \Rightarrow x = 3 \)3. \( 2x-4 = 0 \Rightarrow x = 2 \) Therefore, the critical points are \( x = 1 \), \( x = 3 \), and \( x = 2 \).

Step 5: Verify the Critical Points

Confirm that these critical points are correct within the domain of the function. The derivative calculated \( g'(x) = 0 \) at \( x = 1, 2, \text{ and } 3 \), shows that these are indeed the locations where the slope of the tangent to the function is zero.

Key Concepts

First Derivative

In calculus, the first derivative of a function gives us vital information about the function's behavior. It tells us how the function's value changes as the input changes, essentially giving the rate of change or the slope of the tangent to the curve. For any function, if you want to find where it reaches a peak or a valley (known as critical points), the first step is to determine its first derivative.In our given problem, we need to find the derivative of the function \( g(x) = (x-1)^2(x-3)^2 \). Here, the function is composed of a product of two expressions. Follow the rules of differentiation to find its derivative:- Identify given expressions as \( u(x) = (x-1)^2 \) and \( v(x) = (x-3)^2 \).- Use the product rule to find the derivative: if a function is the product of two other functions, like \( u(x) \) and \( v(x) \), its derivative \( g'(x) \) is \( u'(x)v(x) + u(x)v'(x) \).- Calculate \( u'(x) \) and \( v'(x) \), which are \( 2(x-1) \) and \( 2(x-3) \) respectively.Thus, finding the first derivative is a crucial step that requires understanding of rules like the product rule, which leads us into our next topic.

Product Rule

The product rule is a key principle in calculus used for finding the derivative of a product of two functions. When dealing with polynomial functions, the product rule is especially handy in simplifying expressions.For a function given as a product, say \( g(x) = u(x)v(x) \), the derivative \( g'(x) \) can be found using the formula: \[ g'(x) = u'(x)v(x) + u(x)v'(x) \].Let's break down what we did in the solution:

• We identified \( u(x) = (x-1)^2 \) and \( v(x) = (x-3)^2 \).
• Found \( u'(x) \) as \( 2(x-1) \) and \( v'(x) \) as \( 2(x-3) \), by applying the basic power rule of differentiation.
• Applied the product rule formula to combine these results, resulting in \( g'(x) = 2(x-1)(x-3)^2 + (x-1)^2 \, 2(x-3) \).

Understanding and utilizing the product rule allows us to differentiate complex polynomial expressions efficiently, paving the way to quickly identify critical points in functions.

Polynomial Functions

Polynomial functions form the backbone of many calculus problems, like the one we are solving. These functions are composed of terms involving powers of variables, usually adding constant coefficients to produce a form like \( a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 \).The function we're exploring, \( g(x) = (x-1)^2(x-3)^2 \), is a polynomial because it can be expanded into a series of terms involving powers of \( x \). To find its critical points, we:

• Calculated its first derivative using the product rule.
• Simplified the derivative to solve the equation \( g'(x) = 0 \).
• Identified the values where the function's slope is zero, giving us the critical points \( x = 1 \), \( x = 2 \), and \( x = 3 \).

Polynomial functions often have multiple critical points, which we determine by examining where the derivative equals zero. These critical points are crucial for understanding where a polynomial function may have maximum or minimum values, key insights for graphing and analysis.