Problem 2 - Calculus for Biology and Medicine

Question

\mathrm{\\{} I n ~ P r o b l e m s ~ , each function is continuous and defined on a closed interval. It therefore satisfies the assumptions of the extreme- value theorem. With the help of a graphing calculator or spreadsheet, graph each function and locate its global extrema. (Note that a function may have more than one global minimum or maximum point.) $$ f(x)=-x^{2}+1,-1 \leq x \leq 1 $$

Step-by-Step Solution

Answer

Global maximum at (0, 1); global minima at (-1, 0) and (1, 0).

Step 1: Understand the Function and Interval

The function given is \( f(x) = -x^2 + 1 \). It is a quadratic function with a leading coefficient that is negative, indicating that it opens downwards. The interval provided is \([-1, 1]\), meaning we're only evaluating the function on this domain.

Step 2: Identify the Vertex (Turning Point)

For a quadratic function \( f(x) = ax^2 + bx + c \), the vertex form reveals that the vertex occurs at \( x = -\frac{b}{2a} \). Here, \( a = -1 \), \( b = 0 \), so the vertex is at \( x = 0 \). Substitute into the function: \( f(0) = -(0)^2 + 1 = 1 \). Thus, the vertex is at \( (0, 1) \).

Step 3: Evaluate Endpoints

Evaluate the function at the endpoints of the interval to check for global extrema. At \( x = -1 \), \( f(-1) = -(-1)^2 + 1 = 0 \). At \( x = 1 \), \( f(1) = -(1)^2 + 1 = 0 \).

Step 4: Determine Global Extrema

Compare the values obtained from the vertex and the endpoints. The highest value is \( f(0) = 1 \), which is the global maximum. The lowest values are \( f(-1) = 0 \) and \( f(1) = 0 \), which are the global minima. Thus, the global maximum is at \( (0, 1) \) and global minima are at \( (-1, 0) \) and \( (1, 0) \).

Key Concepts

Understanding Quadratic Functions

A quadratic function is a polynomial function of degree 2, generally represented as \( f(x) = ax^2 + bx + c \).
It has a parabolic graph shape, either opening upwards or downwards, depending on the sign of the leading coefficient \( a \).

• If \( a > 0 \), the parabola opens upwards, often resembling a smile.
• If \( a < 0 \), the parabola opens downwards, looking like a frown.

The function \( f(x) = -x^2 + 1 \) is a specific example of a quadratic function where the parabola opens downwards over the interval of \([-1, 1]\).
This is because the coefficient \( a = -1 \), indicating a downward opening.

Exploring Global Extrema

Global extrema refer to the highest and lowest points on a function over a certain interval.
The extreme-value theorem asserts that if a function is continuous over a closed interval, it will have both a maximum and minimum value in that interval.

• The maximum value is known as the global maximum.
• The minimum value is known as the global minimum.

In the function \( f(x) = -x^2 + 1 \), the global maximum is located at the vertex \((0, 1)\).
This occurs because the vertex is the peak point of the downward-opening parabola.Additionally, the global minima occur at the endpoints of the interval \((-1, 0)\) and \((1, 0)\), where the function takes the lowest values.

Using a Graphing Calculator

A graphing calculator is a valuable tool for visualizing complex functions like quadratic functions.
It allows students to quickly plot functions and observe their behaviors over specific intervals.

• You can enter the function \( f(x) = -x^2 + 1 \) into the calculator.
• Setting the x-values to the desired interval, such as \([-1, 1]\), helps in closely examining the function's behavior.
• Observe the graph to identify the global maximum and minima clearly.

By using a graphing calculator, you can also verify calculations and ensure that your identified extrema align with visual data, providing an excellent confirmation tool in mathematical studies.