Problem 58 - 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). $$ y=\left\\{\begin{array}{ll}{-\frac{1}{4} x^{2}-\frac{1}{2} x+\frac{15}{4},} & {x \leq 1} \\ {x^{3}-6 x^{2}+8 x,} & {x>1}\end{array}\right. $$

Step-by-Step Solution

Answer

Locals maxima: \(y = 3\) at \(x = 1\) and \(y = 4\) at \(x = 4\).

Step 1: Identify the Domain Endpoints

The function is defined piecewise with two parts. For the domain, the endpoint where the function changes is at \(x = 1\). \(f(x)\) is defined for all real numbers, so there are no inherent restrictions from the function expressions themselves.

Step 2: Calculate the Critical Points for Each Piece

To find the critical points, we need the derivative of each piece of the function. For \(x \leq 1\), the derivative is \(y' = -\frac{1}{2}x - \frac{1}{2}\). Setting this equal to zero results in critical points, but since \(-\frac{1}{2}x - \frac{1}{2}\) does not equal zero within \(x \leq 1\), there are no critical points here other than the endpoint. For \(x > 1\), the derivative is \(y' = 3x^2 - 12x + 8\). Setting this equal to zero gives us: \[ 3x^2 - 12x + 8 = 0 \] Solving this quadratic equation by factorization or the quadratic formula gives critical points at \(x = \frac{2}{3}\) and \(x = 4\), but only \(x = 4\) is relevant since \(x > 1\).

Step 3: Evaluate the Function at the Critical Points and Domain Endpoints

Evaluate the function at the identified points: 1. Endpoint \(x = 1\), using \(-\frac{1}{4} x^{2}-\frac{1}{2} x+\frac{15}{4}\), gives \(y = 3\).2. Critical point \(x = 4\), using \(x^{3}-6 x^{2}+8 x\), gives \(y = 4\).Check the behavior at the boundary \(x = 1\) to ensure the function continues smoothly to \(x > 1\) as \( y(1^+) = 3\).

Step 4: Determine the Extreme Values

Compare the function values obtained at the critical points and endpoint to find extreme values. - At \(x = 1\), \(y = 3\) is a local maximum for \(x \leq 1\). - At \(x = 4\), \(y = 4\) serves as a local maximum within \(x > 1\).There are no absolute extreme values since the function is unbounded as \(x \to -\infty \) and \(x \to +\infty \).

Key Concepts

Domain Endpoints

In any mathematical function, domain endpoints are crucial to understanding the scope of the function. They represent the values where the function begins or ends. For piecewise functions like the one given, domain endpoints are the points where the function's behavior may change.
This function has a domain endpoint at \(x = 1\). It's the transition point between the two expressions defining the piecewise function. When \(x \leq 1\), the function is defined by the quadratic expression \(-\frac{1}{4} x^{2}-\frac{1}{2} x+\frac{15}{4}\). On changing to \(x > 1\), the function switches to a cubic expression \(x^{3}-6 x^{2}+8 x\).
Understanding endpoints helps in evaluating the function's continuity and ensuring that it behaves as expected at these junctions. In this case, even though the expressions change at \(x = 1\), the evaluations at this point must remain consistent to guarantee the function's continuity.

Extreme Values

Extreme values in a function refer to the highest or lowest points the function can reach, known as maxima and minima. Finding these values involves analyzing at critical points and domain endpoints.
We observed two important points for this function: \(x = 1\) and \(x = 4\). Upon evaluating these:

• At \(x = 1\), the function value is \(y = 3\). It's considered a local maximum for \(x \leq 1\).
• At \(x = 4\), the function value reaches \(y = 4\), marking it as a local maximum for \(x > 1\).

Though these points give us local extrema, note that this specific function lacks absolute extreme values because it is unbounded, meaning it extends indefinitely in both the negative and positive directions of the \(x\)-axis.

Piecewise Functions

A piecewise function is a function composed of multiple sub-functions, each defined over a certain interval. This type of function is unique because it can exhibit different behaviors or expressions in different parts of its domain.
For this particular exercise, the piecewise function is expressed as:

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

As \(x\) shifts from one interval to the next, the governing rule or expression changes. This requires separate analysis for each piece when seeking critical points or evaluating behaviors. Examining how these parts connect at their domain endpoints ensures a seamless transition, which is imperative for the function's continuity and integration.

Derivatives

Derivatives are fundamental in calculus for determining how a function changes at any point, which brings us to the core of finding critical points.
By taking the derivative of a function, we can identify points where the slope is zero, indicating potential maxima, minima, or saddle points. In this exercise, derivatives help analyze each piece of the piecewise function:

• For \(x \leq 1\), the derivative is \(y' = -\frac{1}{2}x - \frac{1}{2}\). Solving \(y' = 0\) indicates no critical points aside from considering \(x = 1\).
• For \(x > 1\), the derivative \(y' = 3x^2 - 12x + 8\) revealed critical points at \(x = \frac{2}{3}\) and \(x = 4\). Only \(x = 4\) is relevant for this interval, contributing to identifying local extrema.

Derivatives illuminate where the function potentially shifts from increasing to decreasing or vice versa, and provide insight into the overall geometry of the function's graph.