Problem 25 - Thomas Calculus in SI Units

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 at \( (0.5, -4) \) and the absolute maximum is at \( (2, -0.25) \).

Step 1: Determine the derivative of the function

To find critical points, we first need to calculate the derivative of the function. The given function is \( F(x) = -\frac{1}{x^2} \). The derivative of \( F(x) \) can be found using the power rule, where the derivative of \( x^{-n} \) is \( -n \cdot x^{-n-1} \). Therefore, the derivative is:\[ F'(x) = 2x^{-3} = \frac{2}{x^3} \]

Step 2: Find critical points

Critical points occur where the derivative is zero or undefined. Since \( \frac{2}{x^3} \) is never zero, we only need to look for points where the derivative does not exist. This happens when \( x = 0 \), but this is outside our interval \( 0.5 \leq x \leq 2 \). Hence, there are no critical points from the derivative analysis.

Step 3: Evaluate the endpoints of the interval

Since there are no critical points in the interval, we only need to evaluate the function at the endpoints of the interval to find the absolute maximum and minimum values. The endpoints of the interval are \( x = 0.5 \) and \( x = 2 \).Evaluate the function at these points:- \( F(0.5) = -\frac{1}{(0.5)^2} = -\frac{1}{0.25} = -4 \)- \( F(2) = -\frac{1}{2^2} = -\frac{1}{4} = -0.25 \).

Step 4: Identify absolute extrema

The potential absolute extrema are at the values from Step 3:- \( F(0.5) = -4 \) is the lowest value, so it is the absolute minimum.- \( F(2) = -0.25 \) is the highest value, so it is the absolute maximum.

Step 5: Graph the function

To represent the function graphically on the interval \( 0.5 \leq x \leq 2 \), plot the function \( F(x) = -\frac{1}{x^2} \). Identify that the graph is decreasing over this interval, given the nature of the inverse-square function. The coordinates of the extrema found are:- Minimum at \( (0.5, -4) \)- Maximum at \( (2, -0.25) \)

Key Concepts

Absolute Maximum

In calculus, the absolute maximum of a function on a given interval is the highest point over the entire range of that interval. This is the largest value that the function can reach. To find the absolute maximum, we look at critical points within the interval and the endpoints of the interval.

• A critical point is where the derivative is zero or undefined.
• The endpoints are the values at the extreme ends of the interval.

For the function \(F(x) = -\frac{1}{x^2}\) on the interval \(0.5 \leq x \leq 2\), we only consider the endpoints since there are no critical points in this range. Upon evaluation:

• \(F(2) = -0.25\) is the absolute maximum, as it's the highest function value in the interval after checking both endpoints.

Absolute Minimum

The absolute minimum of a function occurs at the point with the lowest function value in the interval.Finding it involves the same steps as determining the absolute maximum:

• Evaluate the function at critical points and endpoints.
• Compare these values to determine which is the smallest.

In our example, \(F(x) = -\frac{1}{x^2}\), we evaluated at the interval's endpoints because there were no critical points:

• The lowest value found was \(F(0.5) = -4\), making it the absolute minimum.

This is the lowest point the function reaches on this interval.

Derivative

A derivative provides the slope of the tangent to the curve at a given point in a function. It indicates how the function is changing at any particular point. To find critical points, which are potential absolute maxima or minima:

• Calculate the derivative of the function.
• Set it to zero to find where the slope is flat.
• Identify where the derivative is undefined to locate more critical points.

For \(F(x) = -\frac{1}{x^2}\), the derivative was found using the power rule:

• \(F'(x) = \frac{2}{x^3}\)
• This expression is never zero within the interval, hence no critical points from zeros of the derivative occur.
• However, \(F'(x)\) is undefined when \(x = 0\), which doesn't affect our interval \([0.5, 2]\).

Graph of Function

Graphing a function helps visualize its behavior across an interval.For \(F(x) = -\frac{1}{x^2}\), the graph on \([0.5, 2]\) can give insights into the characteristics of the function:

• The curve is decreasing throughout the interval.
• The graph shows a steep descent as \(x\) approaches 0.5.
• The absolute minimum occurs at \(x = 0.5\) with a value of \(-4\).
• The absolute maximum is at \(x = 2\) and is \(-0.25\).

Pointing out extrema on the graph helps in identifying key features of the function and understanding where it increases or decreases.