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: Understand the Problem

To find the number that exceeds its square by the maximum amount, we need to find the maximum value of the expression \(x - x^2\) for \(x\) in the interval \([0,1]\).

Step 2: Setup the Function

Define the function \(f(x) = x - x^2\). This function represents the difference between the number \(x\) and its square \(x^2\).

Step 3: Find the Derivative

Calculate the derivative of the function \(f(x) = x - x^2\). The derivative is \(f'(x) = 1 - 2x\). This will help identify where the function reaches its maximum.

Step 4: Find Critical Points

Set the derivative equal to zero and solve for \(x\): \[ 1 - 2x = 0 \]\[ 2x = 1 \]\[ x = \frac{1}{2} \]The critical point of the function is \(x = \frac{1}{2}\).

Step 5: Evaluate the Endpoints and Critical Points

Evaluate the function \(f(x) = x - x^2\) at the critical point and endpoints (0 and 1): - \(f(0) = 0 - 0^2 = 0\)- \(f(1) = 1 - 1^2 = 0\)- \(f\left(\frac{1}{2}\right) = \frac{1}{2} - \left(\frac{1}{2}\right)^2 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}\) The maximum value of the function is \(\frac{1}{4}\), occurring at \(x = \frac{1}{2}\).

Step 6: Conclusion

Since the function \(x - x^2\) is maximized at \(x = \frac{1}{2}\), the number that exceeds its square by the maximum amount on the interval \([0,1]\) is \(\frac{1}{2}\).

Key Concepts

Derivatives

Derivatives play a crucial role in calculus. They help us understand how a function changes at different points in its domain. In this exercise, we have the function \( f(x) = x - x^2 \). To find its derivative, we differentiate the function with respect to \( x \), giving us \( f'(x) = 1 - 2x \). This derivative tells us the slope of the tangent line to the graph of \( f(x) \) at any point \( x \).

• A positive derivative indicates the function is increasing.
• A negative derivative suggests the function is decreasing.
• If the derivative is zero, like when \( f'(x) = 1 - 2x = 0 \), it signifies a potential turning point.

Understanding the derivative is essential for determining where the function might reach its highest or lowest values in a given interval.

Critical Points

Critical points are places on the graph of a function where the derivative is zero or undefined. They are crucial because they indicate where the function's rate of change is racing towards zero, possibly indicating a peak or a valley. For the function \( f(x) = x - x^2 \), setting \( f'(x) = 0 \) gives us \( 1 - 2x = 0 \), leading to \( x = \frac{1}{2} \). This \( x \) value is a critical point.
Critical points can potentially be:

• A local maximum, where the function changes from increasing to decreasing.
• A local minimum, where the function changes from decreasing to increasing.
• A point of inflection, where the function changes concavity.

For our purpose, identifying and evaluating critical points helps us to locate maximum values within an interval.

Maximizing Functions

Maximizing a function means finding the maximum value a function can reach within its domain. In this exercise, we aim to maximize \( f(x) = x - x^2 \) over the interval \([0,1]\). By evaluating the critical point \( x = \frac{1}{2} \) and the endpoints, we can determine which gives the highest value.

• At \( x = 0 \), \( f(0) = 0 \).
• At \( x = 1 \), \( f(1) = 0 \).
• At \( x = \frac{1}{2} \), \( f\left(\frac{1}{2}\right) = \frac{1}{4} \).

So the maximum value is \( \frac{1}{4} \), and it occurs at \( x = \frac{1}{2} \). To maximize a function is about determining where it reaches its peak, considering both critical points and endpoints.

Interval Analysis

Interval analysis involves examining a function's behavior over a specific range of values. In this exercise, we look at the interval \([0,1]\) to find where our function exceeds its square by the maximum amount. We start by identifying and evaluating any critical points and endpoints, as these are potential points for maxima or minima.
In this particular case:

• Determine endpoints: the function begins at \( x = 0 \) and ends at \( x = 1 \)
• Assess the critical point \( x = \frac{1}{2} \)

By evaluating the function at these points within the interval, we determine that the maximum value is \( \frac{1}{4} \) at \( x = \frac{1}{2} \). Interval analysis helps ensure we find the global maximum within a specific range, instead of simply identifying local behavior.