Problem 48 - University Calculus: Early Transcendentals

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 \(x = 1\) and \(x = 3\).

Step 1: Differentiate the Function

To find the critical points, we need to first find the derivative of the function, \( g(x) = (x-1)^2(x-3)^2 \). Let's use the product rule for differentiation. Let \( u = (x-1)^2 \) and \( v = (x-3)^2 \). Then, \( u' = 2(x-1) \) and \( v' = 2(x-3) \). Apply the product rule:\[g'(x) = u'v + uv' = 2(x-1)(x-3)^2 + (x-1)^2(2(x-3))\]

Step 2: Simplify the Derivative

Expand and simplify the expression for \( g'(x) \):\[g'(x) = 2(x-1)(x^2 - 6x + 9) + 2(x^2 - 2x + 1)(x-3)\]Distribute and combine like terms:\[g'(x) = 2((x-1)(x^2 - 6x + 9)) + 2((x^2 - 2x + 1)(x-3))\]Simplified further, this becomes:\[g'(x) = 4x^3 - 16x^2 + 20x - 12\]

Step 3: Find Critical Points

Critical points occur where \( g'(x) = 0 \) or is undefined. Since \( g'(x) \) is a polynomial, it is defined everywhere. Therefore, equate it to zero:\[4x^3 - 16x^2 + 20x - 12 = 0\]Factor out the equation to find the roots.First factor out the greatest common factor of 4:\[x^3 - 4x^2 + 5x - 3 = 0\]This polynomial can factor further or roots can be estimated using various algebraic methods. Completing the factorizing, we find:\[(x-1)^2(x-3) = 0\]

Step 4: Solve for Critical Points

From the factored form \((x-1)^2(x-3) = 0\), solve for \(x\):1. \((x-1)^2 = 0\) gives \(x = 1\)2. \(x-3 = 0\) gives \(x = 3\)The critical points are \(x = 1\) and \(x = 3\).

Key Concepts

Derivative

The derivative of a function is a fundamental concept in calculus, representing the rate at which a function changes at any given point. Imagine you are driving a car; the speed of the car at each moment is akin to the derivative of your journey. It tells you how fast the position changes over time. Mathematically, the derivative of a function at a specific point provides the slope of the tangent line to the graph of the function at that point.

To find critical points, derivatives play a pivotal role as they allow us to determine where the function does not change, i.e., its rate of change is zero. These are the points where the graph might have peaks, valleys, or plateaus. In our exercise, the function is differentiated to find such critical points where the slope of the tangent is zero. This is achieved by setting the derivative to zero and solving for the variable to find the exact points where this occurs.

Product Rule

The product rule is a technique used in calculus when you need to differentiate a function that is the product of two other functions. Having two moving parts, it's like trying to understand how the total stock price moves when both the number of shares and the price per share change. The product rule provides a neat formula for this purpose.

• If you have a function expressed as the product of two functions, say u(x) and v(x), the derivative of this product is found by:
  \[ (g(x))' = u'(x) \, v(x) + u(x) \, v'(x) \]

In our original exercise with the function \( g(x) = (x-1)^2(x-3)^2 \), we let \( u(x) = (x-1)^2 \) and \( v(x) = (x-3)^2 \).

Applying the product rule helps determine how each part of the product contributes to the derivative. By calculating the derivatives \( u'(x) \) and \( v'(x) \), and using the product rule, we extend our tools to effectively find the critical points of the function by further algebraic simplification.

Polynomial

Polynomials form the backbone of many mathematical analyses. They are expressions made up of variables raised to whole number powers and multiplied by coefficients. Think of them as a series of ramps and valleys that can stretch endlessly or fold sharply depending on their degree and coefficients.

• Polynomials are continuous and smooth, meaning they have no breaks or sharp points.
• The degree tells us the highest power of the variable present, which often indicates the maximum number of roots or zeros it might have.

In the exercise, the function \( g(x) = (x-1)^2(x-3)^2 \) simplifies to a polynomial expression for ease of finding the derivative.

As polynomials are differentiable over their domain, finding the derivative of such a function allows us to explore its critical points by setting the first derivative to zero. This approach helps determine the polynomial's peaks (maxima), troughs (minima), or saddle points, facilitating comprehensive analyses of graphs and real-world phenomena.