Problem 56 - Thomas Calculus

Question

In Exercises \(51-58\) , find the critical points and domain endpoints for each function. Then find the value of the function at each of these points and identify extreme values (absolute and local). $$ y=\left\\{\begin{array}{ll}{3-x,} & {x<0} \\ {3+2 x-x^{2},} & {x \geq 0}\end{array}\right. $$

Step-by-Step Solution

Answer

Critical point is at \(x = 1\). Local minimum: \(y(0) = 3\). Local maximum and absolute maximum: \(y(1) = 4\).

Step 1: Identify Critical Points

To find critical points, we start by finding the derivative of each piece of the piecewise function. For the first piece (\(3-x\)), the derivative is \(-1\). For the second piece \(3 + 2x - x^2\), the derivative is \(2 - 2x\). Set the derivative equal to zero to find critical points. \[\frac{dy}{dx} = 2 - 2x = 0\]Solving for \(x\) gives \(x = 1\). Thus, \(x = 1\) is a critical point.

Step 2: Define the Domain and Its Endpoints

The function is defined in two domains: \(x < 0\) and \(x \geq 0\). The domain endpoint for the left piece is \(x \to 0^{-}\) while for the right piece it is \(x \to 0^{+}\). Additionally, for the right piece, it extends to infinity, but the meaningful change happens around critical points and small negative/positive values of \(x\).

Step 3: Evaluate Function at Critical Points and Endpoints

Evaluate the function at the critical point \(x = 1\):\[y(1) = 3 + 2(1) - 1^2 = 4\]Evaluate the function at the endpoint \(x = 0\) for continuity and possible errors:For \(x \to 0^{-}\), \(y = 3 - 0 = 3\).For \(x \to 0^{+}\), \(y = 3 + 2(0) - 0^2 = 3\).Thus, both sides agree at \(x = 0\) with value \(3\).

Step 4: Identify Extremes

Compare the values found at critical point and endpoints of the domain. We have \(y(1) = 4\) and \(y(0) = 3\). These values show that \(y(0) = 3\) is a local minimum while \(y(1) = 4\) is a local maximum. Since this function is continuous and piecewise defined without practical limits, \(y(1) = 4\) is considered an absolute maximum within the given domain segments.

Key Concepts

Piecewise Functions

A piecewise function is a mathematical expression that is defined by different rules for different intervals of its domain. It's like having a function that's divided into pieces, with each piece being applicable over a particular subdomain. For example, in our exercise, the function is defined as:

• For \(x < 0\), the function is \(3 - x\)
• For \(x \geq 0\), the function changes to \(3 + 2x - x^2\)

This means that the behavior of the function changes depending on whether \(x\) is less than zero or zero and above. Understanding piecewise functions is crucial because they often appear in real-world scenarios where conditions change. To solve problems involving piecewise functions, you need to carefully examine the function's definition over each interval and ensure you consider the transitions between these pieces.

Derivative

The derivative of a function represents its rate of change—how the function's output changes as its input changes. When dealing with piecewise functions, we need to find the derivative for each separate piece. In the original exercise:

• The derivative of \(3 - x\) for \(x < 0\) is simply \(-1\), which means the function decreases at a constant rate.
• The derivative of \(3 + 2x - x^2\) for \(x \geq 0\) is \(2-2x\).

To find critical points, where the function could have extreme values, we set the derivative equal to zero. Solving \(2 - 2x = 0\) gives us \(x = 1\). This indicates that at \(x = 1\), the function may have a local maximum or minimum, which is determined further through evaluation. Knowing the derivative helps us understand the shape and direction of the function's graph.

Extreme Values

Extreme values refer to the highest or lowest points that a function achieves within a given domain. These points can be absolute (overall highest or lowest) or local (highest or lowest within a small surrounding area). In the provided exercise:

• The function was evaluated at the critical point \(x = 1\), yielding \(y(1) = 4\). This indicates a local maximum since the derivative changes sign around this point.
• The endpoint \(x = 0\) was also considered, giving \(y(0) = 3\), representing a local minimum, as it is lower than values for small positive \(x\).

Evaluating these points is important as it helps understand where a function reaches its peak or trough within the domain and can provide essential insights in optimization problems and real-world applications.

Domain Endpoints

Domain endpoints are the bounds or edges of the interval over which a function is defined. For piecewise functions, these are critical because they mark where one piece ends and another begins. In the given function:

• The left side endpoint is approached as \(x \to 0^{-}\).
• The right side endpoint starts at \(x \to 0^{+}\) and extends indefinitely since \(x \geq 0\) has no upper bound.

Understanding endpoints is crucial because they can sometimes be where extreme values occur, as seen when there are no other critical points in an interval. Evaluating the function at endpoints helps ensure continuity and can uncover potential discrepancies where pieces of the function meet. They are essential considerations in understanding the full behavior of piecewise functions.