Problem 33 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the formula for the amount \(E(x)\) by which a number \(x\) exceeds its square. Plot a graph of \(E(x)\) for \(0 \leq x \leq 1\). Use the graph to estimate the positive number less than or equal to 1 that exceeds its square by the maximum amount.

Step-by-Step Solution

Answer

The number is \(\frac{1}{2}\) with maximum exceed of \(\frac{1}{4}\).

Step 1: Understanding the Problem

We need to find a formula for the amount by which a number \(x\) exceeds its square. This means we are looking for \(E(x) = x - x^2\). Our task is to represent \(E(x)\) as a function of \(x\) and use a graph to find the maximum value of \(E(x)\) for \(0 \leq x \leq 1\).

Step 2: Derive the Formula

We start with \(x\) and subtract its square: \(E(x) = x - x^2\). This is a quadratic equation that will describe how much \(x\) exceeds \(x^2\).

Step 3: Determine the Range of Interest

Since we are looking at the interval \(0 \leq x \leq 1\), \(E(x)\) can take values in this range. Our goal is to find the point where \(E(x)\) reaches a maximum on this interval.

Step 4: Analyze the Function Behavior

The function \(E(x) = x - x^2\) is a downward-opening parabola because the coefficient of \(x^2\) is negative. The vertex of this parabola will give us the maximum value of \(E(x)\) within \(0 \leq x \leq 1\).

Step 5: Find the Vertex of the Parabola

The vertex of a parabola given by \(ax^2 + bx + c\) is found at \(x = -\frac{b}{2a}\). Here, \(a = -1\) and \(b = 1\), so the vertex is at \(x = -\frac{1}{2(-1)} = \frac{1}{2}\).

Step 6: Evaluate \(E(x)\) at the Vertex

Plug \(x = \frac{1}{2}\) into \(E(x) = x - x^2\) to find maximum exceed amount: \[E\left(\frac{1}{2}\right) = \frac{1}{2} - \left(\frac{1}{2}\right)^2 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}\].

Step 7: Graph the Function

Sketch the graph of \(E(x) = x - x^2\) from \(x = 0\) to \(x = 1\). You'll see a parabola that peaks at \(x = \frac{1}{2}\), with a height of \(\frac{1}{4}\).

Step 8: Conclusion from the Graph

Using the graph, confirm that the maximum value of \(E(x)\) occurs at \(x = \frac{1}{2}\), where it reaches the maximum exceeding value of \(\frac{1}{4}\).

Key Concepts

Parabolas

Parabolas are U-shaped curves that feature prominently in algebra and calculus, particularly when dealing with quadratic functions. The standard equation of a parabola is given by the expression \( y = ax^2 + bx + c \). One key characteristic of a parabola is the direction it opens:

• If the coefficient \( a \) is positive, the parabola opens upwards.
• If \( a \) is negative, the parabola opens downwards.

These shapes are not just visually appealing; they represent the graphical solutions to quadratic equations.
When plotting a quadratic function like \( E(x) = x - x^2 \), you should expect a parabola that opens downwards because the coefficient of \( x^2 \) is negative (-1). This specific feature indicates that the parabola has a maximum point, which is crucial in various applications, including maximizing areas, profits, or other values described by quadratic relationships.

Vertex of a Parabola

The vertex of a parabola is either the highest or lowest point on the graph, depending on the opening direction. For a parabola given by the equation \( ax^2 + bx + c \), the vertex can be found using the formula \( x = -\frac{b}{2a} \).
This formula gives you the x-coordinate of the vertex, after which you can find the y-coordinate by substituting back into the equation.
Let's take the function \( E(x) = x - x^2 \). Here, the coefficients are \( a = -1 \) and \( b = 1 \). Thus, the x-coordinate of the vertex is \( x = -\frac{1}{2(-1)} = \frac{1}{2} \).
By substituting \( x = \frac{1}{2} \) into \( E(x) \), we solve for the maximum y-coordinate and find \( E\left(\frac{1}{2}\right) = \frac{1}{2} - \left(\frac{1}{2}\right)^2 = \frac{1}{4} \).

• This tells us the point (\( \frac{1}{2}, \frac{1}{4} \)) is the vertex of the parabola.
• It's the maximum point for the given quadratic function, which is key for optimization problems.

Optimization in Calculus

Optimization is a fundamental concept in calculus, focusing on finding the maximum or minimum value of a function. It has real-world applications in economics, engineering, and natural sciences, often involving maximizing or minimizing something like cost or profit.
In problems involving quadratic functions, the solution often revolves around finding the vertex of the parabola representing the function.
For the function \( E(x) = x - x^2 \), determining the vertex ensures we find the maximum amount by which a number exceeds its square in the interval \( 0 \leq x \leq 1 \).

• The vertex method simplifies cases where calculus techniques, like using derivatives, may also be applied.
• This specific problem is about locating the point where the parabolic function \( E(x) \) reaches its peak within the specified interval.

Using calculus, one could also use the first and second derivative tests to confirm the vertex as a maximum or minimum point, providing a robust framework for solving a wide range of optimization problems.