Problem 18 - Thomas Calculus

Question

In Exercises \(15-30\) , find the absolute maximum and minimum values of each function on the given interval. Then graph the function. Identify the points on the graph where the absolute extrema occur, and include their coordinates. $$ f(x)=4-x^{2}, \quad-3 \leq x \leq 1 $$

Step-by-Step Solution

Answer

Absolute maximum: 4 at \( x = 0 \); Absolute minimum: -5 at \( x = -3 \).

Step 1: Understand the Problem

We need to find the absolute maximum and minimum values of the function \( f(x) = 4 - x^2 \) on the interval \(-3 \leq x \leq 1\). Then, we'll graph the function and identify the points where the extrema occur.

Step 2: Calculate Critical Points

To find extrema, we first find the derivative of the function \( f(x) = 4 - x^2 \). The derivative is \( f'(x) = -2x \). We set \( f'(x) = 0 \) to find critical points: \( -2x = 0 \). Solving gives \( x = 0 \) as a critical point.

Step 3: Evaluate the Function at Critical Points and Endpoints

We evaluate \( f(x) \) at the critical point and at the endpoints of the interval. Calculate \( f(-3) = 4 - (-3)^2 = 4 - 9 = -5 \). Calculate \( f(0) = 4 - 0^2 = 4 \). Calculate \( f(1) = 4 - 1^2 = 3 \).

Step 4: Identify Absolute Extrema

Compare the function values from Step 3: \( f(-3) = -5 \), \( f(0) = 4 \), \( f(1) = 3 \). The absolute maximum value is 4 at \( x = 0 \), and the absolute minimum value is -5 at \( x = -3 \).

Step 5: Graph the Function and Identify Extrema Points

Graph the function \( f(x) = 4 - x^2 \) over the interval \(-3 \leq x \leq 1\). Mark the points \((0, 4)\) and \((-3, -5)\) on the graph as these are the points where the absolute extrema occur.

Key Concepts

Absolute Extrema

In calculus, absolute extrema refer to the highest or lowest point on a graph over a specified interval. These points are called the absolute maximum and minimum values of the function. When tasked with finding absolute extrema, you are essentially looking for the points where the function reaches its peak and its lowest dip in a given domain.
To determine these points on the graph of a function like \(f(x) = 4 - x^2\) within the interval \([-3, 1]\), begin by identifying both critical points and endpoints as potential candidates for these absolute values. Absolute maxima and minima are often located at critical points, where the derivative of the function equals zero, or at the endpoints of the interval. Evaluating the function at these points gives insight into which nodes produce the extreme values.
In our specific example with \(f(x) = 4 - x^2\), the derivative is \(f'(x) = -2x\) and it is zero at \(x = 0\). Checking this critical point along with endpoints \(x = -3\) and \(x = 1\), we find that the absolute maximum is \(f(0) = 4\) and the absolute minimum is \(f(-3) = -5\).

Critical Points

Critical points are essential in finding where the slope of a function is zero or undefined, signifying potential local maxima or minima. These points occur when the derivative of a function equals zero or does not exist.
For the function \(f(x) = 4 - x^2\), we find the derivative to be \(f'(x) = -2x\). Setting this equal to zero, \(-2x = 0\), we solve for \(x\) and find that \(x = 0\) is a critical point. This critical point could be a location for a local maximum or minimum.
Once you've found the critical points, evaluating the function at these points and comparing them with the function's values at the interval's endpoints leads to identifying the absolute extrema. This process demonstrates why critical points are a crucial step in the calculus of finding extrema.

Graphing Functions

Graphing functions helps visually represent the behavior of equations over a specified interval, including highlighting points of interest such as absolute extrema. By drawing the function \(f(x) = 4 - x^2\), you get a visual depiction of how the function behaves from \(x = -3\) to \(x = 1\).
The graph of \(f(x) = 4 - x^2\) is a downward-opening parabola, indicating the existence of a peak at \(x = 0\), which corresponds with the highest function value in our interval. When graphing, mark key coordinate points like \((0, 4)\) for the absolute maximum and \((-3, -5)\) for the absolute minimum.
By understanding the shape and symmetry of parabolas, one can also predict where these points are likely to lie. The visual representation aids in solidifying your findings from algebraic calculations, ensuring that your determination of extrema is accurate and complete.