Problem 120 - Calculus Volume 1

Question

For the following exercises, find the local and/or absolute maxima for the functions over the specified domain. $$y=\left(x-x^{2}\right)^{2} \text { over }[-1,1]$$

Step-by-Step Solution

Answer

The absolute maximum is 4 at \( x = -1 \).

Step 1: Find the derivative

First, find the derivative of the function to determine critical points. We have the function \( y = (x - x^2)^2 \). Start by using the chain rule and power rule to differentiate. Let \( u = x - x^2 \), so \( y = u^2 \). Then \( \frac{dy}{dx} = 2u \cdot \frac{du}{dx} \). Since \( \frac{du}{dx} = 1 - 2x \), the derivative is \( \frac{dy}{dx} = 2(x - x^2)(1 - 2x) \). Simplify this expression to evaluate it further for critical points.

Step 2: Simplify the derivative

Simplify the expression \( 2(x - x^2)(1 - 2x) \) obtained from the derivative. Expanding the terms results in: \( 2(x - x^2)(1 - 2x) = 2(x - x^2 - 2x^2 + 2x^3) = 2x - 2x^2 - 4x^2 + 4x^3 \), which simplifies to \( 2x - 6x^2 + 4x^3 \). Therefore, \( \frac{dy}{dx} = 2x - 6x^2 + 4x^3 \).

Step 3: Find critical points

Set the derivative \( 2x - 6x^2 + 4x^3 \) equal to zero to find the critical points. Factor this expression: \( 2x(1 - 3x + 2x^2) = 0 \). We can factor further: \( 2x(x - 1)(2x - 1) = 0 \). Therefore, the critical points are \( x = 0 \), \( x = 1 \), and \( x = \frac{1}{2} \).

Step 4: Evaluate the function at critical points and endpoints

Evaluate the original function \( y = (x - x^2)^2 \) at the critical points \( x = 0 \), \( x = 1 \), and \( x = \frac{1}{2} \), as well as at the boundaries of the domain \( x = -1 \) and \( x = 1 \). For \( x = 0 \), \( y(0) = 0^2 = 0 \). For \( x = 1 \), \( y(1) = (1-1)^2 = 0 \). For \( x = \frac{1}{2} \), \( y\left(\frac{1}{2}\right) = \left(\frac{1}{2} - \left(\frac{1}{2}\right)^2\right)^2 = \left(\frac{1}{2} - \frac{1}{4}\right)^2 = \left(\frac{1}{4}\right)^2 = \frac{1}{16} \). At \( x = -1 \), \( y(-1) = [(-1) - (-1)^2]^2 = (-1 - 1)^2 = (-2)^2 = 4 \).

Step 5: Identify maxima

Compare the values obtained from Step 4. The function values are \( y(0) = 0 \), \( y(1) = 0 \), \( y\left(\frac{1}{2}\right) = \frac{1}{16} \), and \( y(-1) = 4 \). The highest value is \( y(-1) = 4 \). So, the absolute maximum over the interval \([-1, 1]\) is 4 at \( x = -1 \).

Key Concepts

calculus

Calculus is a branch of mathematics that delves into the study of change. It combines algebra, geometry, and mathematical analysis to explore rates of change and accumulations. In this exercise, we delve into finding local and absolute maxima.
One key aspect is understanding how functions behave graphically over a given domain. By examining these behaviors, students can assess areas where a function reaches its highest value (maxima) or lowest value (minima). This process involves computing derivatives to pinpoint changes in direction, and evaluating specific points to confirm maxima or minima.
Calculus equips students with the tools to solve real-world problems involving change and optimization, from physics to economics.

critical points

When analyzing a function, finding critical points is essential to understanding its behavior. Critical points are points on a graph where the derivative is zero or undefined. They are potential candidates for local maxima or minima, where the slope of the tangent to the function is flat.
To find critical points, follow these steps:

• Differentiate the function to find the derivative.
• Set the derivative equal to zero and solve for the variable, typically denoted as \( x \).

Critical points may indicate where the function changes direction, which can be a peak or a trough on its graph. Evaluating these points helps determine if they correspond to maxima, minima, or saddle points.

derivative

The derivative is a fundamental concept in calculus, representing the rate of change or slope of a function at any given point. In this exercise, taking the derivative is crucial for locating critical points.
The process involves applying rules like the power rule, product rule, or chain rule depending on the complexity of the function involved. By differentiating the function \( y = (x - x^2)^2 \), we used the chain and product rules.

• Chain Rule: useful for differentiating functions of functions.
• Product Rule: useful when two functions are multiplied together.

Finding derivatives helps uncover where functions ascend or descend, which is vital for identifying maxima and minima.

function evaluation

Function evaluation involves calculating the function's value at specific points. This exercise highlights evaluating both critical points and endpoints of a domain. Local maxima or minima are crucial in optimizing a function's output.
Here's how you evaluate a function:

• Substitute the \( x \)-values of interest (like critical points and boundary points) into the original function.
• Compute the resultant \( y \)-value for each input.

For instance, in the given problem, we evaluate the function \( y = (x - x^2)^2 \) at the points \( x = 0 \), \( x = 1 \), \( x = \frac{1}{2} \), and the boundaries \( x = -1 \) and \( x = 1 \). Comparing these outputs reveals the highest function value, enabling determination of the absolute maximum. This step ensures comprehensive understanding of the function's behavior over its domain.