Problem 142 - Calculus Volume 1

Question

Determine all points on the graph of \(f(x)=x^{3}+x^{2}-x-1\) for which the slope of the tangent line is a. horizontal b. \(-1\)

Step-by-Step Solution

Answer

For horizontal tangents: \( \left(\frac{1}{3}, -\frac{32}{27}\right), (-1, 0) \). For slope \(-1\): \( (0, -1), \left(-\frac{2}{3}, -\frac{5}{27}\right) \).

Step 1: Find the derivative of the function

To find the slope of the tangent line at any point on the graph of the function, we first need to find the derivative of the function. Given the function \( f(x) = x^3 + x^2 - x - 1 \), apply the power rule to derive it. The derivative is:\[ f'(x) = 3x^2 + 2x - 1 \]

Step 2: Find points where the slope is horizontal

A horizontal tangent line means the slope is zero. Set the derivative equal to zero to find where the slope is horizontal:\[ 3x^2 + 2x - 1 = 0 \]Solve this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3 \), \( b = 2 \), and \( c = -1 \).

Step 3: Solve the quadratic equation

Calculate the discriminant \( b^2 - 4ac = 2^2 - 4 \times 3 \times (-1) = 4 + 12 = 16 \). Since the discriminant is positive, there are two real solutions:\[ x = \frac{-2 \pm \sqrt{16}}{6} \]\[ x = \frac{-2 + 4}{6} = \frac{1}{3} \text{ and } x = \frac{-2 - 4}{6} = -1 \].

Step 4: Determine corresponding \( y \)-coordinates for horizontal slope

Substitute \( x = \frac{1}{3} \) and \( x = -1 \) back into the original function to find the \( y \)-coordinates. For \( x = \frac{1}{3} \):\[ f\left(\frac{1}{3}\right) = \left(\frac{1}{3}\right)^3 + \left(\frac{1}{3}\right)^2 - \frac{1}{3} - 1 \]\[ = \frac{1}{27} + \frac{1}{9} - \frac{1}{3} - 1 \]\[ = \frac{1}{27} + \frac{3}{27} - \frac{9}{27} - \frac{27}{27} \]\[ = -\frac{32}{27} \]For \( x = -1 \):\[ f(-1) = (-1)^3 + (-1)^2 + 1 - 1 = -1 + 1 + 1 - 1 = 0 \].The points are \( \left(\frac{1}{3}, -\frac{32}{27}\right) \) and \( (-1, 0) \).

Step 5: Find points where the slope is -1

Set the derivative equal to \(-1\) to find where the slope is \(-1\):\[ 3x^2 + 2x - 1 = -1 \]\[ 3x^2 + 2x = 0 \]Factor the equation:\[ x(3x + 2) = 0 \]Solve for \( x \), giving \( x = 0 \) or \( 3x + 2 = 0 \). Solving \( 3x + 2 = 0 \) gives \( x = -\frac{2}{3} \).

Step 6: Determine corresponding \( y \)-coordinates for slope -1

Substitute \( x = 0 \) and \( x = -\frac{2}{3} \) into the original function to find the \( y \)-coordinates. For \( x = 0 \):\[ f(0) = 0^3 + 0^2 - 0 - 1 = -1 \].For \( x = -\frac{2}{3} \):\[ f\left(-\frac{2}{3}\right) = \left(-\frac{2}{3}\right)^3 + \left(-\frac{2}{3}\right)^2 + \frac{2}{3} - 1 \]\[ = -\frac{8}{27} + \frac{4}{9} + \frac{2}{3} - 1 \]\[ = -\frac{8}{27} + \frac{12}{27} + \frac{18}{27} - \frac{27}{27} \]\[ = -\frac{5}{27} \].The points are \( (0, -1) \) and \( \left(-\frac{2}{3}, -\frac{5}{27}\right) \).

Key Concepts

Tangent Line

A tangent line is a straight line that touches a curve at a single point, without crossing over it. It's like skimming the surface of the curve and aligns itself with the direction the curve is going at that very point. The most interesting aspect of a tangent line in calculus is that its slope at that point corresponds to the derivative of the function evaluated there. This slope gives us the instantaneous rate of change of the function at that point.
To find where the tangent line is horizontal, we need the derivative of the function to equal zero. This is because a horizontal line has a slope of zero.

• The slope of the tangent is found using the derivative.
• A tangent is horizontal when the derivative is zero.

Understanding the concept of a tangent line is crucial because it connects the geometric view with the mathematical, offering insight into how smooth or steep the function is at any given point.

Quadratic Formula

Quadratic equations often arise when solving for the derivative's critical points. The quadratic formula \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]is a reliable tool for finding the roots of any quadratic equation of the form \( ax^2 + bx + c = 0 \).
Using the quadratic formula requires identifying the coefficients \( a \), \( b \), and \( c \) from the derivative set to zero, signifying critical points where the slope changes.

• Identify coefficients \( a \), \( b \), \( c \).
• Calculate discriminant \( b^2 - 4ac \).
• Apply the formula for real roots.

This method is pivotal in calculus problems dealing with derivatives because it helps locate points where the behavior of the tangent line changes, giving clear insight into the nature of the function's graph.

Slope of the Tangent Line

The slope of a tangent line to a function at a particular point is critical for understanding the behavior of that function at the point. This slope is essentially the derivative at that point. The derivative function tells us the slope of the tangent line for any \( x \).
To find exact slopes like horizontal or specific slopes such as -1:

• Set derivative equal to desired slope (e.g., 0 for horizontal).
• Solve the resulting equation for \( x \).

For example, in the context of the function given, solving \( f'(x) = 0 \) helped find horizontal slopes. Similarly, solving \( f'(x) = -1 \) found where the slope -1 occurs. Understanding and calculating the slope of the tangent line is crucial, as it summarizes how steep or shallow the curve is at a specific point.

Function Graph Analysis

The analysis of a function's graph involves understanding its basic shape and the critical features that influence that shape. By using derivatives, you can determine key points that affect how the graph behaves:

• Find where the slope of the tangent changes.
• Identify maximum and minimum points.
• Understand concavity and inflection points.

For the function \( f(x) = x^3 + x^2 - x - 1 \), we calculated points where the tangent was horizontal or had a slope of -1. These critical points gave us locations where the graph was flat and where it tilted downward with a slope of -1.
Graph analysis ensures we see how the graph rises and falls over its domain, enabling us to draw more insightful conclusions about the function's behavior across different values of \( x \). This deeper understanding is vital for applications ranging from physics to economics.