Problem 64

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

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Answer

Critical point: $x = 1$. Absolute maximum: 4 at $x = 1$. Absolute minimum: 3 (at $x = 0$ and as $x \to 0^-$).

1Step 1: Identify the Domain

Identify the conditions under which each piece of the piecewise function is defined. For the piecewise function $ y = \begin{cases} 3-x, & x < 0 \ 3+2x-x^2, & x \geq 0 \end{cases} $, the domains are: $ x < 0 $ for $ y = 3-x $ and $ x \geq 0 $ for $ y = 3 + 2x - x^2 $. Thus, the overall domain of $ y $ is all real numbers.

2Step 2: Find Derivatives

Calculate the derivative of each piece within its respective domain. For $ y = 3-x $, the derivative $ \frac{dy}{dx} = -1 $. For $ y = 3+2x-x^2 $, the derivative $ \frac{dy}{dx} = 2 - 2x $.

3Step 3: Solve for Critical Points

Critical points occur where the derivative is zero or undefined. For $ y = 3-x $, derivative is constant $-1$, so no critical points there. For $ y = 3+2x-x^2 $, set $ \frac{dy}{dx} = 0 $, which is $ 2 - 2x = 0 $. Solving gives $ x = 1 $. Thus, a critical point is at $ x = 1 $.

4Step 4: Evaluate Function at Critical Points and Endpoints

Evaluate the function at critical points and domain boundaries. For $ x < 0 $ boundary, check at $ x = 0^{-} $ ,$ y = 3 - 0 = 3 $. For $ x \geq 0 $ boundary at $ x = 0 $, $ y = 3 + 0 - 0^2 = 3 $, and at critical point $ x = 1 $, $ y = 3 + 2(1) - (1)^2 = 4 $.

5Step 5: Determine Extreme Values

Record the function values at critical points and boundaries: $ y(0^{-}) = 3 $, $ y(0) = 3 $, $ y(1) = 4 $. Therefore, the absolute maximum value of the function is 4 at $ x = 1 $, and the absolute minimum is 3, occurring at $ x = 0 $ and the left endpoint as $ x \to 0^- $. There are no absolute extrema away from these points.

Key Concepts

Critical PointsDerivativesExtreme ValuesFunction Domain

Critical Points

Critical points are specific values of the independent variable where the function's derivative is zero or undefined. These points are crucial because they indicate where the function might have a local maximum, minimum, or a saddle point. In piecewise functions, finding critical points involves calculating the derivative of each piece separately.
In our function, it consists of two pieces:

First piece: For $ x < 0 $, the function is $ y = 3 - x $. Here, the derivative $ \frac{dy}{dx} = -1 $ doesn't equal zero, so there are no critical points in this section.
Second piece: For $ x \geq 0 $, the function is $ y = 3 + 2x - x^2 $. The derivative is $ \frac{dy}{dx} = 2 - 2x $. Set $ \frac{dy}{dx} = 0 $, giving us $ 2 - 2x = 0 $, which solves to $ x = 1 $. Thus, $ x = 1 $ is a critical point in this section.

Finding the critical points is an essential step in analyzing the overall behavior of the function.

Derivatives

Derivatives provide the rate of change of a function with respect to its independent variable. In simpler terms, a derivative tells you how a function's output value changes as its input value changes. Calculating derivatives is key to understanding where a function increases or decreases, and for identifying critical points.
For our piecewise function:

First piece: $ y = 3 - x $, the derivative is $ \frac{dy}{dx} = -1 $. This derivative is constant and negative, indicating a linear decrease in this section.
Second piece: $ y = 3 + 2x - x^2 $, the derivative is $ \frac{dy}{dx} = 2 - 2x $. This is a linear equation where the slope's value depends on $ x $, showcasing a more variable rate of change than the first piece.

By examining the derivatives, you can understand how different sections of the function behave in terms of growth and decline.

Extreme Values

Extreme values, both absolute and local, represent the particular points where a function reaches its highest or lowest values within a given range.
Use the critical points and the boundaries of the domain to find these values. In our case:

Evaluating at the critical point $ x = 1 $, $ y = 3 + 2(1) - (1)^2 = 4 $ which is the function's absolute maximum value.
Check the boundaries: Before $ x = 0 $, $ x = 0^- $ with $ y = 3 $, and at $ x = 0 $, $ y = 3 + 0 - 0^2 = 3 $. These points are the absolute minimum since the lowest value here is 3.

By using the critical points and endpoints, you can effectively identify the highest and lowest points that the function can reach.

Function Domain

The domain of a function is the complete set of values that the function's independent variable, typically $ x $, can take. For piecewise functions, the domain is the combination of each piece's individual domains.
In our example:

The first piece $ y = 3 - x $ is defined for $ x < 0 $.
The second piece $ y = 3 + 2x - x^2 $ is defined for $ x \geq 0 $.

So the overall domain of this piecewise function is all real numbers, from negative infinity to positive infinity. This domain ensures every possible $ x $ value has a corresponding $ y $ value in one of the function's segments. Understanding the domain helps you know where to evaluate critical points and look for extreme values.

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