Problem 20 - 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. $$ F(x)=-\frac{1}{x}, \quad-2 \leq x \leq-1 $$

Step-by-Step Solution

Answer

Maximum value is 1 at \(x = -1\); minimum value is \(\frac{1}{2}\) at \(x = -2\).

Step 1: Determine Critical Points

To find the critical points, we first find the derivative of the function. The function is \( F(x) = -\frac{1}{x} \). Its derivative is \( F'(x) = \frac{1}{x^2} \). Since \( F'(x) \) never equals zero and is undefined only if \( x = 0 \), which is not in our interval, there are no critical points from the derivative.

Step 2: Evaluate Function at Endpoints

Since there are no critical points in the interval, we evaluate \( F(x) \) at the endpoints of the interval: \( x = -2 \) and \( x = -1 \). - Compute \( F(-2) = -\frac{1}{-2} = \frac{1}{2} \)- Compute \( F(-1) = -\frac{1}{-1} = 1 \)

Step 3: Identify Absolute Maximum and Minimum

Compare the function values at the endpoints.- \( F(-2) = \frac{1}{2} \)- \( F(-1) = 1 \)Therefore, the absolute maximum value is 1 at \( x = -1 \), and the absolute minimum value is \( \frac{1}{2} \) at \( x = -2 \).

Step 4: Graph the Function and Plot Extrema

Graph the function \( F(x) = -\frac{1}{x} \) over the interval \([-2, -1]\). Plot the points \((-2, \frac{1}{2})\) and \((-1, 1)\) on the graph.These points \((-2, \frac{1}{2})\) and \((-1, 1)\) correspond to the absolute minimum and maximum, respectively.

Key Concepts

Extrema

In calculus, extrema refer to the maximum and minimum values that a function can have over a certain interval. When we look for extrema, we differentiate between absolute (or global) extrema and relative (or local) extrema. Absolute extrema are the highest or lowest points over the entire interval being considered, while relative extrema are simply the highest or lowest points in a small section around a particular point. In the given exercise, since we evaluate only the interval from \[-2, -1\], the focus is on finding the absolute extrema, which are presented by checking the endpoints for the highest and lowest function values. This process helps us understand how the function behaves across the specified range.

Critical Points

Critical points play a fundamental role when determining extrema. These are the points where the first derivative of a function is either zero or undefined. However, to be an extremum, a critical point must still reside within the domain of the function. For many problems, critical points can guide us in determining where a function may peak or dip.

• If the derivative at a point is zero, it may correspond to a flat tangent, suggesting a peak or valley.
• If the derivative is undefined, the function might have a cusp or vertical tangent.

In the problem at hand, when calculating the derivative of the function, \( F(x) = -\frac{1}{x} \), we found \( F'(x) = \frac{1}{x^2} \). Since this derivative doesn't equal zero anywhere between \(-2\) and \(-1\), and since it is only undefined outside the interval, critical points don't aid us in this instance. Instead, the analysis rests on endpoints.

Derivative

The derivative of a function provides a powerful tool for understanding its behavior. It represents how the function's output value changes as we make small changes in the input value. Specifically, the derivative \( F'(x) \) tells how steep the function \( F(x) \) is at any point \( x \).

• If the derivative is positive, the function is increasing.
• If the derivative is negative, the function is decreasing.
• If the derivative is zero, the function is flat (neither increasing nor decreasing).

In our exercise, the derivative \( F'(x) = \frac{1}{x^2} \) tells us the function is always increasing in the interval since \( \frac{1}{x^2} \) is always positive for non-zero \( x \). Since the derivative never equals zero on the given interval, this confirms there are no critical points driving the extrema within the interval.

Interval Evaluation

Evaluating a function over a specific interval involves calculating the function's value at certain points within that interval, often the endpoints, to determine characteristics like extrema. In scenarios where critical points are absent or don't provide insights within an interval, endpoints become crucial in this assessment.

• To determine the absolute extrema on \([-2, -1]\) for \( F(x) = -\frac{1}{x} \), the function values at \( x = -2 \) and \( x = -1 \) are computed.
• These computations give \( F(-2) = \frac{1}{2} \) and \( F(-1) = 1 \), respectively.

The largest value \( 1 \) at \( x = -1 \) signifies the absolute maximum, while the smallest value \( \frac{1}{2} \) at \( x = -2 \) denotes the absolute minimum. This evaluation ensures we have captured all potential extrema efficiently on the specified interval.