Problem 7 - Calculus Early Transcendentals: Pearson New International Edition

Question

What number exceeds its square by the maximum amount? Begin by convincing yourself that this number is on the interval \([0,1]\).

Step-by-Step Solution

Answer

The number is \( \frac{1}{2} \).

Step 1: Understanding the Problem

We need to find the number that exceeds its square by the maximum amount. This means we should maximize the function given by the expression \[ f(x) = x - x^2 \] for values of \( x \) in the interval \([0, 1]\).

Step 2: Deriving the Function

The expression we need to maximize is given by \[ f(x) = x - x^2 \].This function represents the amount by which a number exceeds its square.

Step 3: Finding the Derivative

To find the critical points where the function might achieve its maximum, we need to find the derivative. The derivative of the function \( f(x) = x - x^2 \) is\[ f'(x) = 1 - 2x \].

Step 4: Setting the Derivative to Zero

To find the critical points, set the derivative equal to zero:\[ 1 - 2x = 0 \].

Step 5: Solving for Critical Points

Solve the equation \( 1 - 2x = 0 \), which gives:\[ 2x = 1 \]\[ x = \frac{1}{2} \].This shows that the critical point is \( x = \frac{1}{2} \).

Step 6: Evaluate the Function at the Boundary and Critical Points

Evaluate the function at the critical point and the boundaries of the interval to determine the maximum value:- \( f(0) = 0 - 0^2 = 0 \)- \( f(1) = 1 - 1^2 = 0 \)- \( f\left( \frac{1}{2} \right) = \frac{1}{2} - \left( \frac{1}{2} \right)^2 = \frac{1}{4} \)Thus, the maximum value occurs at \( x = \frac{1}{2} \).

Step 7: Conclusion

The function \( f(x) = x - x^2 \) achieves its maximum at \( x = \frac{1}{2} \) within the interval \([0, 1]\). Hence, the number that exceeds its square by the maximum amount is \( \frac{1}{2} \).

Key Concepts

Critical Points

Critical points are essential when you're trying to find where a function reaches its maximum or minimum values. They are the points where the function's derivative equals zero or is undefined, but primarily we focus on where it equals zero in basic optimization problems. In the case of the function \( f(x) = x - x^2 \), we found the derivative \( f'(x) = 1 - 2x \). By setting this derivative equal to zero, we determined the critical point:

• Set \( 1 - 2x = 0 \)
• Solve for \( x \) yielding \( x = \frac{1}{2} \)

This critical point \( x = \frac{1}{2} \) is where the function is neither increasing nor decreasing, indicating a potential maximum or minimum. Why do we check critical points? Think of them as potential turning points in the story of a function, like peaks or valleys in a landscape. Since our interval was \([0, 1]\), this critical point was a candidate for the maximum value.

Derivative

The derivative of a function gives us insightful information about its behavior. It tells us how the function is changing at any point and allows us to analyze the function's increases and decreases. When we take the derivative of \( f(x) = x - x^2 \), we apply basic rules of calculus to yield:
\[ f'(x) = 1 - 2x \]This derivative formula reflects the rate of change. It's like having a roadmap to understanding the inner workings of the function. When the derivative is positive, the function is rising, and when it’s negative, the function is falling. At \( x = \frac{1}{2} \), the derivative is zero, marking a potential change in direction. In simpler terms, the derivative acts like a guide that tells you if you're climbing uphill or heading downhill on the graph of a function.

Maximizing Functions

Maximizing a function means finding the point or points where it achieves its highest value within a given interval. In optimization problems, once we identify critical points and evaluate the function at these points and boundaries of the interval, we can conclude where the maximum value occurs. For \( f(x) = x - x^2 \), with the critical point at \( x = \frac{1}{2} \), we evaluated:

• \( f(0) = 0 \)
• \( f(1) = 0 \)
• \( f\left( \frac{1}{2} \right) = \frac{1}{4} \)

By comparing these values, it’s evident that the maximum value of \( \frac{1}{4} \) occurs at \( x = \frac{1}{2} \). In everyday life, maximizing functions is like finding the tallest peak on a mountain or the best outcome in a situation. It's all about knowing where the biggest payoff lies.