Problem 29 - Calculus of a Single Variable

Question

Finding Extrema on a Closed Interval In Exercises \(17-36\) , find the absolute extrema of the function on the closed interval. $$ y=3-|t-3|,[-1,5] $$

Step-by-Step Solution

Answer

The absolute maximum of the function \(y=3-|t-3|\) on the interval \([-1,5]\) is 3 at \(t=3\), and the absolute minimum is 1 at \(t=-1\) and \(t=5\).

Step 1: Understand the Function

The function \(y = 3 - |t - 3|\) is equivalent to two functions: \(y = 3 - (t - 3)\) when \(t >= 3\), and \(y = 3 - (3 - t)\) when \(t < 3\).

Step 2: Find Critical Points in Domain

Critical points occur where the derivative of the function is zero or where the derivative does not exist. The derivative of \(3 - t\) is \(-1\) and the derivative of \(3 - (3 - t)\) or \(t\) is \(1\). As neither derivative ever equals zero, there are no critical points due to the derivative. However, the derivative does not exist at \(t = 3\), so \(t = 3\) is the critical point.

Step 3: Evaluate Function at Critical Points and Endpoints

The function should be evaluated at the critical point \(t = 3\) and at the endpoints \(t = -1\) and \(t = 5\) of the interval. For these \(t\) values, the function \(y = 3 - |t - 3|\) equals 3, 1, and 1, respectively.

Step 4: Determine Absolute Extrema

The absolute maximum of the function on the interval is the largest of the function values found in Step 3, which is 3. The absolute minimum is the smallest, which is 1.

Key Concepts

Absolute Extrema

When you are dealing with calculus problems involving extrema, you're essentially identifying the highest and lowest points of a function over a certain range. In this context, absolute extrema refer to:

• Absolute Maximum: The highest point over the entire interval. For the function given, it's 3 at the point \(t = 3\).
• Absolute Minimum: The lowest point over the entire interval. Here, it's 1 at both endpoints \(t = -1\) and \(t = 5\).

These values are crucial in understanding the behavior of a function across a specified domain. We use absolute extrema to assess the full scope of the function's range of output values.

Closed Interval

A closed interval is characterized by including both of its endpoint values in the calculations or consideration for extrema. In mathematical terms, this is denoted with square brackets. For instance, the interval \([-1, 5]\) means that both \(t = -1\) and \(t = 5\) are considered for evaluating extrema.
Closed intervals are significant because they ensure that potential maximums and minimums at the boundaries of a function's domain are not neglected. By assessing the function at these endpoints, we can precisely determine the absolute extrema.

Critical Points

Critical points in calculus are where a function’s derivative is zero or fails to exist, signaling potential changes in extrema.

• Where Derivative Equals Zero: This condition suggests a horizontal tangent to the curve, indicating potential peaks or valleys.
• Where Derivative Fails: The function might have a cusp or sharp point. In our function \(y = 3 - |t-3|\), the critical point is at \(t=3\) where the derivative doesn't exist.

Identifying critical points allows us to accurately pinpoint where changes in increase or decrease might occur, which are essential in finding absolute extrema.

Derivative

The derivative of a function measures how the function's value changes as its input changes; it represents the function's rate of change. In our function, the different pieces have simple linear derivatives.

• For \(y = 3 - (t - 3)\), the derivative is \(-1\).
• For \(y = t\), the derivative is \(1\).

A derivative helps determine where a function is increasing, decreasing, or at a turning point. Calculating derivatives is fundamental as it leads to finding critical points, which then aid in locating absolute extrema on a closed interval.

Piecewise Function

A piecewise function is defined by different expressions based on the input's value. This particular function \(y = 3 - |t - 3|\) becomes piecewise because the absolute value operation defines two different behaviors:

• For \(t >= 3\), the expression is \(y = 3 - (t - 3)\).
• For \(t < 3\), the expression is \(y = 3 - (3 - t)\).

Recognizing a function as piecewise is crucial, especially in finding extrema, because each piece may have different derivatives and therefore different critical points and potential maximums or minimums depending on the domain like the closed interval with endpoints.