Problem 75 - University Calculus: Early Transcendentals

Question

Find the critical points, domain endpoints, and extreme values (absolute and local) for each function. $$y=\left\\{\begin{array}{ll}-x^{2}-2 x+4, & x \leq 1 \\\\-x^{2}+6 x-4, & x>1\end{array}\right.$$

Step-by-Step Solution

Answer

Critical points are \(x = -1\) and \(x = 3\). Absolute maximum: \(y=5\). Local minimum: \(y=1\) at \(x=1\).

Step 1: Identify the Domain of the Function

To find the domain of the piecewise function, review each piece's condition. The first piece, \(-x^2-2x+4\), applies for \(x \leq 1\) and the second piece, \(-x^2+6x-4\), for \(x > 1\). The domain of the entire function is thus all real numbers, \(x \in (-\infty, \infty)\).

Step 2: Find the Derivative of Each Piece

For each piece of the function, differentiate separately:1. For \(-x^2-2x+4\), the derivative is \(f'(x) = -2x - 2\).2. For \(-x^2+6x-4\), the derivative is \(f'(x) = -2x + 6\).

Step 3: Find Critical Points of Each Piece

Critical points occur where the derivative is zero or undefined.1. For \(x \leq 1\), solve \(-2x - 2 = 0\) to get \(x = -1\).2. For \(x > 1\), solve \(-2x + 6 = 0\) to get \(x = 3\).Since \(-1\) is within the domain \(x \leq 1\) and \(3\) is within \(x > 1\), both are critical points.

Step 4: Evaluate Function at Domain Endpoint and Critical Points

Calculate the function's value at the critical points and endpoints of the domain to identify potential extreme values.1. Evaluate \(-x^2 - 2x + 4\) at \(x = 1\): \(y = -1^2 - 2(1) + 4 = 1\).2. Evaluate \(-x^2 - 2x + 4\) at critical point \(x = -1\): \(-(-1)^2 - 2(-1) + 4 = 5\).3. Evaluate \(-x^2 + 6x - 4\) at critical point \(x = 3\): \(-3^2 + 6(3) - 4 = 5\).

Step 5: Determine Extreme Values

Compare computed values to determine extremes:- The value at \(x = -1\) and \(x=3\) both yield \(y = 5\), giving local maxima.- The value at \(x = 1\) yields \(y = 1\), indicating a local minimum.Thus, absolute maximum is \(5\) (at both \(x = -1\) and \(x = 3\)), and there is no absolute minimum in the domain \((-\infty, \infty)\).

Key Concepts

Understanding Piecewise Functions

A piecewise function is a mathematical expression defined by multiple sub-functions, each applied to a specific interval of the domain. This ensures that different conditions or formulas apply based on the value of the input variable.

• The given function is split into two parts:
  • For values where \( x \leq 1 \), the function is \( -x^2 - 2x + 4 \).
  • For values where \( x > 1 \), the function becomes \( -x^2 + 6x - 4 \).
• This setup allows the function to model different behaviors in different intervals.

When dealing with piecewise functions, always pay attention to the given domains. Solving and analyzing these types of functions involves evaluating each piece separately before combining the results to understand the function fully.

Derivative Calculation Basics

The derivative of a function gives us the rate of change at any point on the function. For piecewise functions, we calculate derivatives separately for each piece.

• For the piece \( -x^2 - 2x + 4 \), the derivative is \( f'(x) = -2x - 2 \).
• For the piece \( -x^2 + 6x - 4 \), the derivative is \( f'(x) = -2x + 6 \).

This differentiation enables us to analyze the function's behavior in its respective pieces, especially identifying where the slope is zero. Calculating derivatives precisely involves applying rules like the power rule and combining like terms, often resulting in a linear expression.

Identifying Extreme Values

Extreme values are the highest or lowest points on a function within a certain interval. They help us understand the function's limits in terms of output values.

• To find these, investigate endpoints and critical points.

With our piecewise function:

• We evaluate each sub-function at its endpoint: \( x = 1 \) gives \( y = 1 \)
• At the critical points \( x = -1 \) and \( x = 3 \), \( y = 5 \).

This analysis defines extreme values which include local maxima at \( x = -1 \) and \( x = 3 \), with the function's peaks being at \( y = 5 \). A local minimum is defined at \( x = 1 \) with \( y = 1 \). No absolute minimum exists due to the unbounded domain on either end.

Critical Points Analysis

Critical points help in pinpointing where the function changes direction, i.e., from increasing to decreasing or vice versa. They occur where the derivative is zero or undefined.

• For \( x \leq 1 \), setting the derivative \( -2x - 2 = 0 \) gives a critical point at \( x = -1 \).
• For \( x > 1 \), setting \( -2x + 6 = 0 \) results in \( x = 3 \) as another critical point.

These points are then tested to determine whether they are maxima, minima, or points of inflection by evaluating the function at these points.
After evaluating, they confirm local maxima at these critical points because the function takes the highest value at these points compared to their neighboring intervals.