Problem 65 - College Algebra and Calculus: An Applied Approach

Question

Use a graphing utility to graph the function over the interval. Find the average value of the function over the interval. Then find all \(x\) -values in the interval for which the function is equal to its average value. $$ f(x)=4-x^{2} \quad[-2,2] $$

Step-by-Step Solution

Answer

First, the graph of the function \(f(x) = 4 - x^2\) over the interval [-2, 2] was sketched. The average value of the function over the interval was then calculated using the formula for the average value of a function using integrals and solving it. Finally, the \(x\)-values in the interval for which the function was equal to the average value were obtained by setting the function equal to the average value and solving for \(x\).

Step 1: Graphing the function

First, use a graphing utility to sketch the function \(f(x) = 4 - x^2\). This will help in visualizing the function's behavior over the interval \([-2, 2]\). The function represents a downward-facing parabola with the vertex at (0,4).

Step 2: Calculating the average value

To find the average value of \(f(x)\) over the interval \([-2, 2]\), we can use the formula for the average value of a function on the interval [a, b]:\[\frac{1}{b - a} \int_{a}^{b} f(x) dx \]Here, \(a = -2\), \(b = 2\), \(f(x) = 4 - x^2\). Substituting these values gives:\[\frac{1}{2-(-2)} \int_{-2}^{2}(4-x^2) dx = \frac{1}{4} \int_{-2}^{2}(4-x^2) dx\]Calculate the integral and solve for the average value.

Step 3: Finding \(x\)-values where the function equals the average value

Once the average value is found, set the function \(4-x^2\) equal to this average value and solve for \(x\). These will be the \(x\)-values in the interval \([-2,2]\) where the function \(f(x)\) equals its average value.

Key Concepts

Graphing Functions

Visualizing a function can greatly help in understanding its characteristics and behavior over a specified interval. The function given here, \( f(x) = 4 - x^2 \), is a perfect example of a simple quadratic function that can be graphed easily. When graphing functions, having a graphing utility makes this task straightforward, as it quickly sketches the curve and shows its general shape. In this particular case, the function represents a parabola that opens downward because the coefficient of \( x^2 \) is negative.

The vertex of the parabola is at \( (0, 4) \), which can be observed as the highest point on the graph. Understanding the shape of the graph allows us to see the symmetry and the interval we deal with, which in this case is \([-2, 2]\). This knowledge helps determine integral limits and analyze how the function behaves from left to right along these intervals. Graphs are essential for pinpointing critical points and can also verify solutions or findings from other calculations.

Integrals

Integrals play a crucial role in calculation, especially when determining the average value of a function over a certain interval. The process begins with setting up a definite integral for the function over the defined interval. In this exercise, we aim to find the average value of \( f(x) = 4 - x^2 \) over \([-2, 2]\).

The formula to calculate the average value is:

• \( \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \)

Substituting \( a = -2 \) and \( b = 2 \) into the formula provides \( \frac{1}{4} \int_{-2}^{2} (4 - x^2) \, dx \). Solving this integral involves finding the antiderivative of \( f(x) \) and evaluating it from \(-2\) to \(2\).

Integration gives the total "area" under the curve over this period, and the division by the interval's length adjusts it for calculating the average. By understanding and using integrals, one can not only determine this average but also explore the area under curves that has wide applications in real-life scenarios, like physics and engineering.

Parabolas

A parabola is a fundamental concept in graphing quadratic functions. The function given, \( f(x) = 4 - x^2 \), follows the standard form \( ax^2 + bx + c \) with coefficients \( a = -1 \), \( b = 0 \), and \( c = 4 \). This demonstrates how the parabola has a vertex at \( (0, 4) \) and opens downward due to the negative \( a \) value.

Key characteristics of parabolas include:

• Vertex: the peak or the lowest point depending on the direction of the opening. For this exercise, it is \( (0, 4) \).
• Axis of symmetry: a vertical line through the vertex \( x = 0 \) that divides the parabola into two mirrored halves.
• Direction of opening: determined by the sign of \( a \); negative means downward.

Understanding these properties helps see how parabolas interact with other mathematical concepts like integration. This exercise illustrates finding average values and points where the function equals this average, demonstrating the connection between algebraic rules and visual graphs. Hence, acknowledging parabola characteristics can help students deeply understand both real-world applications and theoretical concepts.