Problem 49 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Let \(f(x)=\frac{1}{4} x^{4}-\frac{1}{3} x^{3}-x^{2} .\) Find the point(s) on the graph of \(f\) where the slope of the tangent line is equal to: a. \(-2 x\) b. 0 c. \(10 x\)

Step-by-Step Solution

Answer

The points where the slope is equal to: a. -2x are \((0, 0)\) and \((1, -\frac{1}{3})\). b. 0 are \((-1, -\frac{5}{12})\), \((0, 0)\), and \((2, -\frac{4}{3})\). c. 10x are \((-3, -\frac{17}{12})\), \((0, 0)\), and \((4, -\frac{32}{3})\).

Step 1: Find the derivative of the function

To find the derivative of the function \(f(x) = \frac{1}{4}x^4 - \frac{1}{3}x^3 - x^2\), use the power rule for derivatives, which states that if \(f(x) = x^n\), then \(f'(x) = nx^{n-1}\). Applying this to each term, we obtain: \(f'(x) = \frac{1}{4}(4x^3) - \frac{1}{3}(3x^2) - (2x)\) Simplify: \(f'(x) = x^3 - x^2 - 2x\)

Step 2a: Find the point(s) where the slope is equal to -2x

Set the derivative equal to -2x: \(x^3 - x^2 - 2x = -2x\) Add 2x to both sides and factor: \(x^3 - x^2 = 0\) \(x^2(x - 1) = 0\) From this, we have two solutions: x = 0 and x = 1. Plug these values back into the original function to find the corresponding y-values: \(f(0) = \frac{1}{4}(0)^4 - \frac{1}{3}(0)^3 - (0)^2 = 0\) \(f(1) = \frac{1}{4}(1)^4 - \frac{1}{3}(1)^3 - (1)^2 = -\frac{1}{3}\) The points where the slope is equal to -2x are: \((0, 0)\) and \((1, -\frac{1}{3})\).

Step 2b: Find the point(s) where the slope is equal to 0

Set the derivative equal to 0: \(x^3 - x^2 - 2x = 0\) Factor out an x: \(x(x^2 - x - 2) = 0\) Factor the quadratic: \(x(x - 2)(x + 1) = 0\) The x-values where the slope is equal to 0 are x = 0, x = -1, and x = 2. Plug these values back into the original function to find the corresponding y-values: \(f(-1) = \frac{1}{4}(-1)^4 - \frac{1}{3}(-1)^3 - (-1)^2 = \frac{1}{4} + \frac{1}{3} - 1 = -\frac{5}{12}\) \(f(0) = 0\) \(f(2) = \frac{1}{4}(2)^4 - \frac{1}{3}(2)^3 - (2)^2 = -\frac{4}{3}\) The points where the slope is equal to 0 are: \((-1, -\frac{5}{12})\), \((0, 0)\), and \((2, -\frac{4}{3})\).

Step 2c: Find the point(s) where the slope is equal to 10x

Set the derivative equal to 10x: \(x^3 - x^2 - 2x = 10x\) Add (2x - 10x) to both sides: \(x^3 - x^2 - 12x = 0\) Factor out an x: \(x(x^2 - x - 12) = 0\) Factor the quadratic: \(x(x - 4)(x + 3) = 0\) The x-values where the slope is equal to 10x are x = 0, x = -3, and x = 4. Plug these values back into the original function to find the corresponding y-values: \(f(-3) = \frac{1}{4}(-3)^4 - \frac{1}{3}(-3)^3 - (-3)^2 = -\frac{17}{12}\) \(f(0) = 0\) \(f(4) = \frac{1}{4}(4)^4 - \frac{1}{3}(4)^3 - (4)^2 = -\frac{32}{3}\) The points where the slope is equal to 10x are: \((-3, -\frac{17}{12})\), \((0, 0)\), and \((4, -\frac{32}{3})\).

Key Concepts

Tangent Line Slope

The slope of a tangent line to a curve at a given point is a crucial concept in calculus. It describes the steepness of the curve precisely at that point. Think of it as the slope you would experience if you were walking along the curve at that exact location.
This slope can tell us a lot about the behavior of the function locally.To find the tangent slope for the function \( f(x) = \frac{1}{4}x^4 - \frac{1}{3}x^3 - x^2 \), we first calculate its derivative. This derivative, \( f'(x) \), is actually the slope of the tangent line at any point \( x \) on the function.

• This derivative is found using the power rule, resulting in: \( f'(x) = x^3 - x^2 - 2x \).
• From this, setting \( f'(x) \) equal to various expressions, like \(-2x\), 0, or \(10x\), reveals the specific points of the graph where these slope values occur.

Understanding tangent line slopes helps us understand instantaneous rates of change, a foundational principle in calculus.

Derivative Calculation

Derivative calculation is a method used to find the rate at which a function changes at any given point. At its core, it involves algebraic manipulations to apply calculus tools, like the power rule.For our function \( f(x) = \frac{1}{4}x^4 - \frac{1}{3}x^3 - x^2 \), the steps are:

• Use the power rule: For any term \( x^n \), the derivative \( f'(x) \) is calculated as \( nx^{n-1} \).
• Each term is processed individually: \( \frac{1}{4}(4x^3), \frac{1}{3}(3x^2), \) and \(2x \).
• Simplifying gives the derivative: \( f'(x) = x^3 - x^2 - 2x \).

This calculated derivative aids in finding critical points where tangent slopes, such as \(-2x\), 0, or \(10x\), equate to linear expressions or constants.

Polynomial Functions

Polynomial functions are a type of mathematical function characterized by terms raised to whole-number exponents. They are essential in various fields, capturing behaviors ranging from simple parabolas to complex curves.Our given function \( f(x) = \frac{1}{4}x^4 - \frac{1}{3}x^3 - x^2 \) is a polynomial.

• The highest exponent (degree) of this function is 4, making it a quartic polynomial.
• Polynomial functions are continuous and smooth, meaning there's no break or sharp point in their graphs.
• Each term's exponent can provide insights into the function's behavior: for instance, as \( x \) moves further from zero, \( x^4 \) will greatly influence the function's value.

By examining polynomial functions through derivatives, we can explore critical points and determine aspects such as peak, trough, or inflection points in the graph.