Problem 25 - University Calculus: Early Transcendentals

Question

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. $$F(x)=-\frac{1}{x^{2}}, \quad 0.5 \leq x \leq 2$$

Step-by-Step Solution

Answer

The absolute minimum is -4 at x = 0.5, and the absolute maximum is -0.25 at x = 2.

Step 1: Find the critical points

To find critical points for the function \[ F(x) = -\frac{1}{x^2}\]we first need to calculate the derivative. Using the power rule, we have \[ F'(x) = \frac{d}{dx}(-x^{-2}) = 2x^{-3} = \frac{2}{x^3}\].Set this derivative equal to zero and solve for x:\[ \frac{2}{x^3} = 0 \].Since the exponential function never equalizes zero, this derivative does not yield any critical points.

Step 2: Evaluate endpoints

The function has no critical points in its domain since the derivative is never zero. Therefore, we evaluate the function at the endpoints of the interval: \( x = 0.5 \) and \( x = 2 \).Evaluate at \( x = 0.5 \):\[ F(0.5) = -\frac{1}{(0.5)^2} = -\frac{1}{0.25} = -4 \].Evaluate at \( x = 2 \):\[ F(2) = -\frac{1}{(2)^2} = -\frac{1}{4} = -0.25 \].

Step 3: Determine absolute extrema

Compare the function values at the endpoints.- At \( x = 0.5 \), \( F(x) = -4 \), the resulting value is the minimum value.- At \( x = 2 \), \( F(x) = -0.25 \), the resulting value is the maximum value.Thus, the absolute maximum is \(-0.25\) at \((x, F(x)) = (2, -0.25)\) and the absolute minimum is \(-4\) at \((x, F(x)) = (0.5, -4)\).

Step 4: Sketch the graph

Plot the curve of \( F(x) = -\frac{1}{x^2} \)on the interval \([0.5, 2]\). Use the evaluated endpoints to identify key points.- The point \((0.5, -4)\) is the lowest point in this interval.- The point \((2, -0.25)\) is the highest point in this interval.Make sure the curve approaches the horizontal asymptote (y = 0) as x increases, showing the decreasing trend as x increases.

Key Concepts

Critical Points

Critical points are values of \(x\) at which the derivative of a function is either zero or undefined. These points can indicate where the function changes direction, which helps in finding local maximum and minimum values (also called extrema).
In the given exercise, the function \(F(x) = -\frac{1}{x^2}\) was considered. The derivative \(F'(x) = \frac{2}{x^3}\) was found using the power rule.
The next step was to set the derivative to zero to find the critical points. However, since \(\frac{2}{x^3} = 0\) results in no solution, there are no critical points within the given interval. It's important to always check the domain as some functions might not have critical points, especially those involving division by \(x\).

Derivative

The derivative of a function represents the rate at which the function's value changes as \(x\) changes. It is essentially the slope of the tangent line at any point on the function's graph.
To find the derivative of \(F(x) = -\frac{1}{x^2}\), we apply the power rule of differentiation. By rewriting the function as \(-x^{-2}\), the derivative becomes \(F'(x) = \frac{2}{x^3}\).
This derivative is critical for analyzing how the function behaves, by providing information on the increasing or decreasing nature of the function over intervals, and helping identify potential locations for extrema.

Extrema

Extrema refer to the maximum or minimum values a function can attain within a given interval. These include both absolute and relative extremum points.
In this exercise, we are interested in determining the absolute extrema of the function \(F(x) = -\frac{1}{x^2}\) on the interval \(0.5 \leq x \leq 2\).
Since there were no critical points found inside the interval, the extrema needed to be evaluated at the endpoints \(x = 0.5\) and \(x = 2\).
This approach is typical in optimization problems when the derivative doesn’t yield any critical changes within the domain.

Absolute Maximum

The absolute maximum of a function on a closed interval is the highest value the function reaches on that interval.
In this scenario, after evaluating the endpoints, it was found that \(F(x) = -0.25\) when \(x = 2\) is the highest value the function attains.
This value is therefore the absolute maximum in this interval. It indicates the peak performance of the function, as no other values on this interval exceed it.

Absolute Minimum

An absolute minimum is the lowest point or value that a function reaches on a particular interval.
For the function \(F(x) = -\frac{1}{x^2}\), calculating the function value at the endpoints revealed \(F(x) = -4\) when \(x = 0.5\) as the absolute minimum.
This result shows where the function dips to its lowest point within the interval. Identifying such points is essential in various applications to understand where the function is at its minimum performance.