Problem 10 - Thomas Calculus

Question

Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. \begin{equation} y=6-2 x-x^{2} \end{equation}

Step-by-Step Solution

Answer

The local and absolute maximum is at (-1, 7); no inflection points exist.

Step 1: Find the First Derivative

To locate the extrema, we begin by finding the first derivative of the given function. The function is \( y = 6 - 2x - x^2 \). Using basic differentiation rules, we find the first derivative: \( y' = \frac{d}{dx}(6 - 2x - x^2) = -2 - 2x \).

Step 2: Critical Points

To find critical points, set the first derivative equal to zero: \(-2 - 2x = 0\). Solving for \(x\), we get \(x = -1\). Substitute \(x = -1\) back into the original function to find the corresponding \(y\)-value: \(y = 6 - 2(-1) - (-1)^2 = 6 + 2 - 1 = 7\). Thus, the critical point is \((-1, 7)\).

Step 3: Determine Local Extrema

Use the second derivative test to determine if the critical point is a local maximum or minimum. First, find the second derivative: \( y'' = \frac{d}{dx}(-2 - 2x) = -2 \). Since the second derivative is negative, it indicates that the critical point \((-1, 7)\) is a local maximum.

Step 4: Absolute Extrema Analysis

As this is a quadratic function (parabola), it opens downwards. Thus, the local maximum at \((-1, 7)\) is also the absolute maximum, as the ends of the parabola will continue to \(-\infty\).

Step 5: Find the Inflection Points

Inflection points occur where the concavity of the function changes, found by setting the second derivative to zero: \(-2 = 0\). Since \(-2\) does not equal zero, there are no inflection points for this function.

Step 6: Graph the Function

Plot the graph of \(y = 6 - 2x - x^2\) using the critical point. It is a downward opening parabola with vertex (local and absolute maximum) at \((-1, 7)\). There are no inflection points.

Key Concepts

Critical Points

Critical points in calculus are where the first derivative of a function is zero or undefined. These points are important because they can show where the function might have local maxima or minima. To find a critical point, we take the derivative of a function and set it equal to zero, solving for the variable. For example, given the quadratic function \( y = 6 - 2x - x^2 \), we find the first derivative: \( y' = -2 - 2x \). Setting \( y' = 0 \) gives \( -2 - 2x = 0 \), leading to \( x = -1 \). Inserting \( x = -1 \) into the original function results in \( y = 7 \), pinpointing the critical point at \((-1, 7)\). Critical points help us in determining the behavior of the function.

Local Extrema

Local extrema refer to the highest or lowest points in a small, localized section of a graph. These can either be a local maximum or local minimum. Once a critical point is established, we perform the second derivative test to conclude if it is an extremum. Given the second derivative \( y'' = -2 \) for our function, it's consistently negative. This implies concavity facing downwards, confirming the critical point as a local maximum. Thus,

• A downward concavity suggests a maximum at that point.
• Negative second derivative means local maximum.

The local extremum for the function \( y = 6 - 2x - x^2 \) is at \((-1, 7)\).

Absolute Extrema

Absolute extrema are the highest or lowest points on the entire function curve, also known as global extrema. For quadratic functions like \( y = 6 - 2x - x^2 \), determining if a point is an absolute extremum is straightforward because it forms a parabola. If the parabola opens downwards which it does here due to the negative coefficient of \( x^2 \), the local maximum is also the absolute maximum since it doesn’t have higher points globally.

• The parabola's vertex represents the highest point.
• No absolute minimum as ends tend to \(-\infty\).

The absolute extrema of the function is located at \((-1, 7)\), making it both a local and absolute maximum.

Inflection Points

Inflection points mark where a function changes its concavity, switching from concave up to concave down, or vice versa. We find these by setting the second derivative to zero or identifying where it doesn't exist. For \( y = 6 - 2x - x^2 \), the second derivative \( y'' = -2 \) shows no change in sin (it’s consistently negative).

• No solution for \( y'' = 0 \) implies no inflection points.
• Absence of sign change indicates maintained concavity.

Thus, there are no inflection points in this quadratic function, confirming a continuous concave-down shape.

Quadratic Function

A quadratic function is of the form \( ax^2 + bx + c \), with its graph forming a parabola. For \( y = 6 - 2x - x^2 \), it reflects a standard quadratic function with

• \(-x^2\) indicating a downward-facing parabola,
• vertex given by solving for critical points.

Such functions are characterized by their symmetry around the vertex. Here, the vertex and hence the critical point \((-1, 7)\) is both a local and absolute maximum. Understanding this helps plot accurate graphs and predict function behavior across the plane.