Problem 74 - University Calculus: Early Transcendentals

Question

Find the critical points, domain endpoints, and extreme values (absolute and local) for each function. $$y=\left\\{\begin{array}{ll} 3-x, & x<0 \\ 3+2 x-x^{2}, & x \geq 0 \end{array}\right.$$

Step-by-Step Solution

Answer

Local max at (1, 4); no absolute ext. values over the domain.

Step 1: Understand the Function

The given function, \( y \), is a piecewise function with two parts: \( y = 3-x \) for \( x < 0 \) and \( y = 3+2x-x^2 \) for \( x \geq 0 \). We will analyze each part separately to find the critical points and extreme values.

Step 2: Determine the Domain

The domain of the function is all real numbers. The piecewise function's definition includes all \( x < 0 \) and all \( x \geq 0 \), collectively covering all real numbers.

Step 3: Find Critical Points for \( x < 0 \)

For the function part \( y = 3-x \), compute the derivative: \( \frac{dy}{dx} = -1 \). Since \( -1 eq 0 \), there are no critical points for \( x < 0 \).

Step 4: Find Critical Points for \( x \geq 0 \)

For \( y = 3+2x-x^2 \), find the derivative: \( \frac{dy}{dx} = 2 - 2x \). Set \( \frac{dy}{dx} = 0 \) to find critical points: \( 2 - 2x = 0 \Rightarrow x = 1 \). Check that \( x = 1 \) is within the domain \( x \geq 0 \).

Step 5: Evaluate the Function at Critical Points

Calculate \( y \) at the critical point \( x = 1 \): \( y = 3 + 2(1) - 1^2 = 4 \). So \( (1, 4) \) is a critical point.

Step 6: Evaluate at Domain Endpoint \( x = 0 \)

For \( x \geq 0 \), evaluate the function at the endpoint \( x = 0 \): \( y = 3 + 2(0) - 0^2 = 3 \). Therefore, \( (0, 3) \) is a relevant point.

Step 7: Evaluate the Function as \( x \rightarrow -\infty \)

For \( x < 0 \), as \( x \rightarrow -\infty \), \( y = 3-x \rightarrow \infty \). This analysis ensures no finite absolute extreme in this section.

Step 8: Determine Extreme Values

By comparing values from Steps 5 and 6, and understanding the behavior as \( x \rightarrow -\infty \):- Local maximum at \( (1, 4) \).- Endpoint value \( (0, 3) \) is neither local max nor min.- No absolute maximum or minimum across the entire piecewise function.

Key Concepts

Piecewise Functions

When dealing with piecewise functions, you should know that these functions are defined by different expressions depending on the value of the independent variable, usually denoted as \( x \). This construction allows the function to express different behaviors for different intervals of \( x \). In the given exercise, the function is defined as \( y = 3-x \) for \( x < 0 \) and \( y = 3+2x-x^2 \) for \( x \geq 0 \).
This means:

• For \( x \) values that are less than 0, the function is linear, showing a constant negative slope of -1.
• For \( x \) values that are 0 or greater, the function presents a quadratic behavior, with a parabolic shape, opening downwards.

Understanding the domains and the nature of the expressions is vital when analyzing piecewise functions.
To study a piecewise function properly, analyze the different pieces individually for their mathematical properties, such as slope, curvature, and point of intersections.

Extreme Values

Finding extreme values involves discovering the highest and lowest points on a graph of a function. These can be local (relative) or absolute. Locating these values is crucial in understanding the limits and behavior of functions. To determine extreme values in piecewise functions, you review each segment separately.
In our example:

• The segment \( y = 3-x \) does not have a peak or valley because it is a straight line continuing infinitely as \( x \) goes to \(-\infty\).
• In the segment \( y = 3+2x-x^2 \), we derived a critical point at \( x=1 \). At this point, the function reaches a local maximum of 4, as the value drops for \( x \) values increasing or decreasing from the point.

Additionally, consider the endpoints of the domain, like \( x=0 \), which evaluates to 3. This does not represent a local extremum but is key in understanding the function's overall range of values.
Always remember, in piecewise contexts, respect the boundaries of each segment, identifying how extreme behavior might shift at transitional points.

Derivative

A derivative measures how a function's output changes concerning changes in its input. It's fundamental in identifying critical points—those spots where a function might achieve extremum values or exhibit inflection.
For the exercise at hand:

• The piece \( y = 3-x \) has a constant derivative of \(-1\). Because this derivative isn't zero, no critical points exist in this interval.
• For \( y = 3+2x-x^2 \), the derivative became \( \frac{dy}{dx} = 2 - 2x \). Setting this derivative to zero, you find \( x=1 \) as a critical point.

A zero derivative implies a horizontal tangent. This suggests the potential for an extremum—something further confirmed by analyzing values around this point.
Understanding derivatives aids in revealing the rates of change, slopes, and resulting shape of graph sections, thus guiding towards the discovery of critical behavioral points.