Problem 11 - Calculus Single Variable

Question

In each of Exercises \(7-22,\) use Fermat's Theorem to locate each \(c\) for which \(f(c)\) is a candidate extreme value of the given function \(f\) $$ f(x)=(x-3)(x+5)^{3} $$

Step-by-Step Solution

Answer

Critical points where extreme values may occur are at \(x = -5\) and \(x = 1\).

Step 1: Identify the Function and Its Type

The function provided is \( f(x) = (x-3)(x+5)^3 \). It's a polynomial function expressed as a product of two factors: a linear factor \((x-3)\) and a cubic factor \((x+5)^3\). To locate critical points, we need to find the derivative and set it equal to zero.

Step 2: Differentiate the Function

We apply the product rule to differentiate the given function. The product rule states that \((uv)' = u'v + uv'\), where \(u = (x-3)\) and \(v = (x+5)^3\). Thus,- \(u' = 1\)- \(v' = 3(x+5)^2\).Therefore, the derivative \(f'(x)\) is:\[ f'(x) = (x-3) \cdot 3(x+5)^2 + (x+5)^3 \cdot 1 \].

Step 3: Simplify the Derivative

Now, we expand and simplify the derivative expression:\[ f'(x) = 3(x-3)(x+5)^2 + (x+5)^3 \]Factor out \((x+5)^2\) from both terms:\[ f'(x) = (x+5)^2 [3(x-3) + (x+5)] \]Simplify inside the brackets:\[ f'(x) = (x+5)^2 (3x - 9 + x + 5) = (x+5)^2 (4x - 4) = 4(x+5)^2 (x-1) \]

Step 4: Find Critical Points

Fermat's Theorem states that critical points occur where \(f'(x) = 0\). Set the expression equal to zero:\[ 4(x+5)^2 (x-1) = 0 \]This equation is satisfied when:- \((x+5)^2 = 0\), giving \(x = -5\)- \(x-1 = 0\), giving \(x = 1\).The critical points are \(x = -5\) and \(x = 1\).

Step 5: Exclusion of Non-Extreme Candidates

Consider the multiplicity of the roots. Since \((x+5)^2\) is a repeated factor, \(x = -5\) might correspond to a point where the function doesn't have an extreme value but rather a more subtle feature like a point of inflection. Meanwhile, \(x = 1\) is a simple factor and thus a candidate for an extreme value.

Key Concepts

Critical Points

Critical points in a function are where something interesting happens, such as a peak or a valley. These occur where the derivative of a function equals zero or is undefined.
They are the potential candidates for finding the highest and lowest points (extreme values) of the function on a graph.

• The nature of these points can indicate local minima, maxima, or even inflection points.
• In our exercise, the critical points are determined by setting the derivative equal to zero (using Fermat's Theorem) and solving for the values of \(x\).

Identifying these points helps us understand where the function changes its increasing or decreasing behavior. It's like finding the spots where you might need to change gears on a winding road.

Derivative

The derivative of a function represents the rate at which the function's value changes. Think of it as the speedometer of a car, showing how fast the position on a graph is changing.

• Finding the derivative is crucial for locating critical points because it tells us the slope of the function at any point.
• In this exercise, we use the derivative to find where the slope is zero, indicating potential extreme values of the polynomial function.
• The derivative is essential because it allows us to find instantaneous rates of change and understand the behavior of functions.

By calculating the derivative in this exercise, we gain insights into the critical points where significant changes in the graph occur.

Polynomial Function

A polynomial function is a mathematical expression involving a sum of powers in one or more variables multiplied by coefficients.
Simply put, these are expressions like \(x^2 + 2x + 1\). In our exercise, the function \(f(x) = (x-3)(x+5)^3\) is a polynomial.

• Polynomials are smooth and continuous, without gaps or jumps.
• The degree of a polynomial tells us the highest power of the variable, which influences its shape and number of turning points.
• In the provided function, the product of a linear and a cubic term indicates a fifth-degree polynomial, accounting for three potential extreme points.

These functions are pivotal in calculus because they offer a simple, yet detailed look into complex behavior through their roots and turning points.

Product Rule

The product rule is a tool used in calculus to find the derivative of a product of two functions. When you have a situation where two functions are multiplied, this rule is indispensable.

• It states: If \(u(x)\) and \(v(x)\) are functions, then the derivative of their product is \((uv)' = u'v + uv'\).
• This is important for accurately differentiating functions like in this exercise where \(f(x) = (x-3)(x+5)^3\).
• The rule helps us break down complex expressions into manageable parts, making it easier to differentiate.

In practical terms, the product rule allows us to understand how two separate rates of change combine into one, giving a clearer picture of the overall change in a polynomial function.