Problem 61 - Applied Calculus

Question

Let \(f(x)=x^{3}-6 x^{2}-15 x+20\). Find \(f^{\prime}(x)\) and all values of \(x\) for which \(f^{\prime}(x)=0 .\) Explain the relationship between these values of \(x\) and the graph of \(f(x)\).

Step-by-Step Solution

Answer

The derivative is \(f'(x) = 3x^2 - 12x - 15\) with critical points at \(x = 5\) and \(x = -1\). These points signify places where the graph's slope is zero.

Step 1: Differentiate the Polynomial

We start by finding the derivative of the function \(f(x) = x^3 - 6x^2 - 15x + 20\). Apply the power rule to differentiate each term. The derivative of \(x^3\) is \(3x^2\), the derivative of \(-6x^2\) is \(-12x\), the derivative of \(-15x\) is \(-15\), and the constant \(20\) becomes zero. Thus, the derivative is \(f'(x) = 3x^2 - 12x - 15\).

Step 2: Set the Derivative Equal to Zero

To find the critical points, set the derivative \(f'(x)\) equal to zero. So we have the equation \(3x^2 - 12x - 15 = 0\). This will help us find the values of \(x\) where the slope of the tangent to the curve is zero, indicating potential local maxima, minima, or points of inflection.

Step 3: Solve the Quadratic Equation

Solve the quadratic equation \(3x^2 - 12x - 15 = 0\). First, simplify by dividing the entire equation by 3 to get \(x^2 - 4x - 5 = 0\). Factor this simplified quadratic to \((x - 5)(x + 1) = 0\). Solving for \(x\), we find \(x = 5\) and \(x = -1\).

Step 4: Analyze the Critical Points on the Graph

The solutions \(x = 5\) and \(x = -1\) are critical points for the function. On the graph of \(f(x)\), these are the points where the slope of the tangent is zero, which means they could be points of local maxima, minima, or points of inflection. To determine the nature of these critical points, further analysis using the second derivative or a sign chart could be performed.

Key Concepts

Derivative

When studying calculus, the concept of a derivative gives us the rate at which a function is changing at any given point. To find the derivative, we use the power rule which involves multiplying the power of each term by its coefficient and then reducing the power by one.
For example, for the function \(f(x) = x^3 - 6x^2 - 15x + 20\), the power rule is applied to yield the derivative \(f'(x) = 3x^2 - 12x - 15\). Each term's power is decreased by one, indicating the change in the slope of the function at various points along its curve. This derivative tells us the slope of \(f(x)\) at any point \(x\).
Being able to find derivatives is fundamental as it sets the stage for analyzing the behavior of functions more deeply.

Critical Points

Critical points occur where the derivative of a function is zero or undefined. Finding these points involves setting the derivative \(f'(x)\) equal to zero and solving for \(x\).
In the example with \(f(x) = x^3 - 6x^2 - 15x + 20\), the derivative is \(f'(x) = 3x^2 - 12x - 15\). Setting this equation equal to zero allows us to solve for critical points:

• Simplify the equation: \(3x^2 - 12x - 15 = 0\).
• Solve by factoring or using the quadratic formula, resulting in \(x = 5\) and \(x = -1\).

These values are critical because they represent the x-values where the slope of the tangent to \(f(x)\) is zero, suggesting a potential local extremum or point of inflection.

Quadratic Equation

Solving quadratic equations is a common requirement in calculus, especially when finding critical points. A quadratic equation can often arise from setting a polynomial's derivative to zero.
For instance, simplifying \(3x^2 - 12x - 15 = 0\) by dividing all terms by 3 results in \(x^2 - 4x - 5 = 0\).

• This can be solved either by factoring: \((x - 5)(x + 1) = 0\),
• or using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

Finding the solutions \(x = 5\) and \(x = -1\) illustrates necessary algebraic skills. These solutions are crucial in calculus, as they can delineate changes in a function's behavior.

Graph Analysis

Graph analysis is crucial to understanding the behavior of a function. It helps in visualizing how the function behaves at its critical points - whether these are peaks, valleys, or points where the function's direction changes.
For the function \(f(x) = x^3 - 6x^2 - 15x + 20\), after computing the derivative and solving for critical points \(x = 5\) and \(x = -1\), we can analyze the graph.

• At each critical point, the slope is zero, suggesting potential local maximums or minimums, or possibly points of inflection.
• To determine their nature, one can perform a further analysis, such as using the second derivative test or a sign chart, to see whether they represent highs, lows, or points of concavity change.

Graph analysis brings the mathematics to life, allowing us to see and interpret the abstract numbers in a visual and more comprehensible form.