Problem 5 - The Calculus with Analytic Geometry

Question

Find the slope of the tangent line to the graph at the point \(\left(x_{1}, y_{1}\right) .\) Make a table of values of \(x, y\), and \(m\) at various points on the graph, and include in the table all points where the graph has a horizontal tangent. Draw a sketch of the graph. $$ y=x^{3}-3 x $$

Step-by-Step Solution

Answer

The slope of the tangent line at point \(a, a^3 - 3a\) is \ 3a^2 - 3. Horizontal tangents occur at \(1, -2\) and \((-1, 2\).

Step 1 - Find the Derivative

To determine the slope of the tangent line, start by finding the derivative of the function. The given function is y = x^3 - 3x. The derivative, y', represents the slope of the tangent line to the graph.Calculate the derivative: \[ y' = \frac{d}{dx} (x^3 - 3x) = 3x^2 - 3 \]

Step 2 - Evaluate the Slope at the Given Point

Choose a point \((x_1, y_1)\) on the graph. Suppose \((x_1, y_1) = (a, a^3 - 3a)\). Substitute x_1 = a into y' to get the slope at the given point:\[ m = y'(a) = 3a^2 - 3 \]

Step 3 - Make a Table of Points and Their Slopes

Generate a table with values of x, y, and the corresponding slope m at various points. Include calculations to find y and y' for different values of x. y = x^3 - 3xy' = 3x^2 - 3 For example:\[\begin{array}{|c|c|c|}\hline x & y & m \ \hline -2 & -2^3 - 3(-2) = -8 + 6 = -2 & 3(-2)^2 - 3 = 12 - 3 = 9 \ -1 & -1^3 - 3(-1) = -1 + 3 = 2 & 3(-1)^2 - 3 = 3 - 3 = 0 \ 0 & 0^3 - 3(0) = 0 & 3(0)^2 - 3 = -3 \ 1 & 1^3 - 3(1) = 1 - 3 = -2 & 3(1)^2 - 3 = 3 - 3 = 0 \ 2 & 2^3 - 3(2) = 8 - 6 = 2 & 3(2)^2 - 3 = 12 - 3 = 9 \ \hline\end{array}\]

Step 4 - Identify Points with Horizontal Tangents

Horizontal tangents occur where the slope, y', is zero. Set the derivative y' equal to zero and solve for x: \[ 3x^2 - 3 = 0 \]\[ 3(x^2 - 1) = 0 \]\[ x^2 - 1 = 0 \]\[ (x - 1)(x + 1) = 0 \]The solutions are x = 1 and x = -1. So, the points with horizontal tangents are (1, -2) and (-1, 2). Add these points to the table.

Step 5 - Draw the Graph

Sketch the graph of y = x^3 - 3x. Mark all critical points, including the points with horizontal tangents found in Step 4. The critical points can help in drawing an accurate curve. The basic shape can be derived by plotting these points: high points, low points, and where the function crosses the x-axis and y-axis.

Key Concepts

derivative

In calculus, the derivative of a function measures the rate at which the function's value changes at a particular point. It is a fundamental tool for finding the slope of tangent lines. For the function given in the exercise, \( y = x^3 - 3x \), the derivative is calculated to find how steep or flat the graph is at any point. The derivative of the function is calculated as follows: \[ y' = \frac{d}{dx} (x^3 - 3x) = 3x^2 - 3 \] This derivative tells us that at any point \( x \), the slope of the tangent line is \( 3x^2 - 3 \).

horizontal tangent

A horizontal tangent line touches the graph of a function without sloping up or down—it is completely flat. In mathematical terms, this happens when the slope (the derivative) of the function is zero. For the given function, horizontal tangents occur where the derivative equals zero: \[ 3x^2 - 3 = 0 \] Solving this equation, we find that the x-values where the tangent is horizontal are \( x = 1 \) and \( x = -1 \). These points correspond to \( (1, -2) \) and \( (-1, 2) \) on the graph. These points are critical because they show where the function changes from increasing to decreasing or vice versa.

critical points

Critical points of a function occur where its derivative is zero or undefined. They are key in understanding the function’s behavior. In the exercise, the points where the derivative \( 3x^2 - 3 \) equals zero (the same points where horizontal tangents are found) are critical points. These points, \( (1, -2) \) and \( (-1, 2) \), are essential because they help identify the extrema (maximum or minimum points) on the graph. To classify these points further (as maxima, minima, or saddle points), one could take the second derivative test, but for this problem, identifying the zero-slope points is enough to understand the shape of the graph better.

graph sketching

Graph sketching involves plotting a function's critical points, understanding its behavior, and drawing an accurate curve based on these observations. Following the steps in the solution, once you have the table of values and critical points, sketching the graph of \( y = x^3 - 3x \) becomes straightforward. Here are some guidelines:

• Plot known points like \( (1, -2) \) and \( (-1, 2) \), where the slope is zero and mark these points distinctly.
• Identify where the function intersects the axes (y-intercept at \( (0,0) \) and check other intersections if any).
• Determine whether the function is increasing or decreasing by examining intervals around critical points.

By incorporating all this information, you achieve a more precise and well-understood sketch of the function’s graph, highlighting high, low, and crossing points effectively.