Problem 6 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 5-26, identify the critical points and find the maximum value and minimum value on the given interval. $$ h(x)=x^{2}+x ; I=[-2,2] $$

Step-by-Step Solution

Answer

Minimum: -1/4 at x=-1/2; Maximum: 6 at x=2.

Step 1: Find the derivative

To find the critical points, first, take the derivative of the function \( h(x) = x^2 + x \). The derivative is found using basic differentiation rules: \[ h'(x) = 2x + 1 \]

Step 2: Set the derivative to zero

Set the derivative equal to zero to find critical points: \[ 2x + 1 = 0 \] Solving for \( x \): \[ 2x = -1 \] \[ x = -\frac{1}{2} \]

Step 3: Evaluate the function at critical points and endpoints

Now evaluate \( h(x) \) at the critical point and the endpoints of the interval \( I = [-2, 2] \): - \( h(-2) = (-2)^2 + (-2) = 4 - 2 = 2 \)- \( h\left(-\frac{1}{2}\right) = \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4} \)- \( h(2) = (2)^2 + 2 = 4 + 2 = 6 \)

Step 4: Determine minimum and maximum values

Compare the values obtained: - \( h(-2) = 2 \)- \( h\left(-\frac{1}{2}\right) = -\frac{1}{4} \)- \( h(2) = 6 \)The minimum value is \(-\frac{1}{4}\) at \( x = -\frac{1}{2} \), and the maximum value is \(6\) at \( x = 2 \).

Key Concepts

Maximum and Minimum Values

When analyzing functions, one common task is to find the maximum and minimum values. These values can tell us a lot about the behavior of the function. In particular, they can help us understand how the function grows and shrinks over a certain interval.

• Maximum Value: The highest point in a specified interval. For example, with the function given by \( h(x) = x^2 + x \), we found that the maximum value is 6 at \( x = 2 \).
• Minimum Value: The lowest point in that interval. In the same function, the minimum value is \( -\frac{1}{4} \) at \( x = -\frac{1}{2} \).

By evaluating these points, we can understand where a function reaches its peaks and troughs. The endpoints of the interval \( I = [-2, 2] \) must also be considered, as maximum or minimum values can occur there as well.

Derivative

A derivative measures how a function changes at any point, essentially giving us the rate of change or the slope of the function at that point. When dealing with polynomial functions like \( h(x) = x^2 + x \), calculating the derivative allows us to identify critical points where maximum or minimum values might occur.To find the derivative, we differentiate the function:

• For \( h(x) = x^2 + x \), the derivative is \( h'(x) = 2x + 1 \).

Setting \( h'(x) = 0 \) is how we locate the critical points. Solving \( 2x + 1 = 0 \) gives us the critical point \( x = -\frac{1}{2} \).Understanding derivatives is crucial in calculus, as they provide insight into the function’s increasing or decreasing behavior, and they pinpoint moments where the direction changes, leading to extreme values (maxima or minima).

Calculus

Calculus is a branch of mathematics that studies continuous change. It consists of two main parts: differentiation, which we've encountered in finding the derivative, and integration. While calculus can seem complex, its basic principles allow us to solve many types of problems in mathematics and applied fields.

• In this exercise, differentiation helped us to understand the behavior of the function \( h(x) = x^2 + x \).
• By determining the critical points, we can assess where the maximum or minimum values occur over a given interval.

This particular example shows how calculus can be applied to real-world problems by finding how a function behaves, peaks, and dips—all essential for optimization problems and more intricate mathematical modeling.