Problem 76 - University Calculus: Early Transcendentals

Question

Find the critical points, domain endpoints, and extreme values (absolute and local) for each function. $$y=\left\\{\begin{array}{ll}-\frac{1}{4} x^{2}-\frac{1}{2} x+\frac{15}{4}, & x \leq 1 \\\x^{3}-6 x^{2}+8 x, & x>1\end{array}\right.$$

Step-by-Step Solution

Answer

Critical points are at \( x = -1, \frac{4}{3}, 2 \); absolute max is 4 at \( x = -1 \); absolute min is 0 at \( x = 2 \).

Step 1: Determine the Domain

The domain of the given piecewise function is all real numbers since there are no restrictions on the values of \( x \) for either piece of the function. The intervals are \( x \leq 1 \) and \( x > 1 \).

Step 2: Find Critical Points for Each Piece

To find critical points, we first need to find the derivative for each part:1. For \( x \leq 1 \), \( y = -\frac{1}{4}x^2 - \frac{1}{2}x + \frac{15}{4} \). The derivative is \( y' = -\frac{1}{2}x - \frac{1}{2} \).2. For \( x > 1 \), \( y = x^3 - 6x^2 + 8x \). The derivative is \( y' = 3x^2 - 12x + 8 \).Set \( y' = 0 \) for each piece and solve for \( x \):- For \( x \leq 1 \), \(-\frac{1}{2}x - \frac{1}{2} = 0 \Rightarrow x = -1 \) (valid as \( x \leq 1 \)).- For \( x > 1 \), \(3x^2 - 12x + 8 = 0 \) gives critical points as \( x = \frac{4}{3}, 2 \). Both are valid since they are greater than 1.

Step 3: Evaluate Function at Critical Points and Domain Endpoints

Substitute the critical points and endpoints into their respective function parts:- At \( x = -1 \), \( y = -\frac{1}{4}(-1)^2 - \frac{1}{2}(-1) + \frac{15}{4} = 4 \).- At \( x = 2 \), \( y = (2)^3 - 6(2)^2 + 8(2) = 0 \).- At endpoint \( x = 1 \), check continuity: - Left side at \( x = 1 \), \( y = -\frac{1}{4}(1)^2 - \frac{1}{2}(1) + \frac{15}{4} = 3 \). - Right side at \( x = 1 \), \( y = (1)^3 - 6(1)^2 + 8(1) = 3 \).Since both parts are equal, \( y = 3 \) is a valid value.

Step 4: Determine the Extreme Values

Evaluate which of these values are local and absolute extremes:- At \( x = -1 \), \( y = 4 \), which is an endpoint for \( x \leq 1 \) and is the highest value reached, making it a local maximum for that interval.- At \( x = 2 \), \( y = 0 \), which is a critical point for \( x > 1 \), is a local minimum since values around it are higher.- At endpoint \( x = -\infty \), the limit goes to infinity since it goes to \( \frac{15}{4} \), and at \( x = \infty \), the endpoint value is also unbounded. - The absolute maximum occurs at \( x = -1 \) with \( y = 4 \); the absolute minimum is \( y = 0 \) at \( x = 2 \).

Step 5: Summary of Critical Points and Extreme Values

Critical points are at \( x = -1 \), \( x = \frac{4}{3} \), and \( x = 2 \). The calculated function values are 4, approximately 2.37, and 0, respectively. The absolute maximum of the function is at \( x = -1 \) with \( y = 4 \) and the absolute minimum is at \( x = 2 \) with \( y = 0 \).

Key Concepts

Domain Endpoints

In calculus, understanding the domain endpoints of a function is essential to evaluate its behavior at the extremes of its domain. A domain endpoint is a value of \( x \) where the piecewise sections of a function begin or end. In our exercise, the domain of the piecewise function is all real numbers. However, it's split into two intervals: \( x \leq 1 \) and \( x > 1 \). The point \( x = 1 \) serves as a critical shift between the two segments. For piecewise functions, checking endpoints like \( x = 1 \) is vital to ensure continuity. This involves calculating the function's value just before and after the endpoint. In our solution, it was shown that for both sides of \( x = 1 \), the evaluated function value is equal, which confirms the function is continuous across this breakpoint.

Piecewise Function

A piecewise function consists of separate expressions defined over particular intervals of the domain. Each piece is governed by its own rule, which makes piecewise functions flexible for modeling different kinds of real-world phenomena. In the original function provided, two distinct expressions cover different segments of the domain:

• \( y = -\frac{1}{4} x^{2} - \frac{1}{2} x + \frac{15}{4} \) for \( x \leq 1 \)
• \( y = x^{3} - 6x^2 + 8x \) for \( x > 1 \)

This allows each part to handle different domain intervals separately, which is why it's essential to analyze them individually. For continuity and derivative calculations, each piece must be considered within its valid range, which can sometimes lead to different critical points and behaviors.

Derivative

The derivative of a function represents the rate at which the function's value changes at any given point, essentially describing the function's slope. For piecewise functions, each piece must have its own derivative evaluated separately. In our situation, the derivatives are

• For \( x \leq 1 \): \( y' = -\frac{1}{2}x - \frac{1}{2} \)
• For \( x > 1 \): \( y' = 3x^2 - 12x + 8 \)

By setting these derivatives to zero, we find the critical points, where the rate of change is zero, potentially indicating local maxima or minima. Critical points are the starting point for determining extreme values, and they help us identify where the function’s slope changes within the given domain.

Extreme Values

Extreme values refer to the highest and lowest points (either absolute or local) a function reaches within its domain. To find them in a piecewise function, the function is evaluated at critical points and domain endpoints. From the step-by-step solution:

• At \( x = -1 \), \( y = 4 \), signifying a local maximum.
• At \( x = 2 \), \( y = 0 \), indicating a local and absolute minimum for \( x > 1 \).
• Both ends at \( x = -\infty \) and \( x = +\infty \) show unbounded behavior asymptotically approaching the limits.

Through this examination, the function's absolute maximum is confirmed at \( x = -1 \) with a value of 4. Identifying these extremes is crucial in understanding the overall behavior and range of the function across its domain.