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Ch. 4 - Applications of Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 4 - Applications of DerivativesProblem 4.1.35
Chapter 4, Problem 4.1.35

Absolute Extrema on Finite Closed Intervals


In Exercises 21–36, 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(t) = 2 − |t|, −1 ≤ t ≤ 3

Verified step by step guidance
1
Step 1: Understand the function f(t) = 2 - |t|. This function involves the absolute value of t, which affects its behavior. The absolute value function |t| is piecewise, meaning it behaves differently based on the sign of t.
Step 2: Break down the function into its piecewise components. For t >= 0, |t| = t, so f(t) = 2 - t. For t < 0, |t| = -t, so f(t) = 2 + t. This gives us two expressions to consider: f(t) = 2 - t for t >= 0 and f(t) = 2 + t for t < 0.
Step 3: Evaluate the function at the endpoints of the interval [-1, 3]. Calculate f(-1) and f(3) using the appropriate piecewise expressions. These values will help determine the extrema.
Step 4: Check for critical points within the interval. Since the function is piecewise linear, it doesn't have any critical points where the derivative is zero. However, the point where the piecewise function changes, t = 0, should be evaluated as it might be an extremum.
Step 5: Compare the values obtained at the endpoints and the point t = 0 to determine the absolute maximum and minimum values. The largest value will be the absolute maximum, and the smallest will be the absolute minimum. Graph the function over the interval [-1, 3] and mark the points where these extrema occur, including their coordinates.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Absolute Extrema

Absolute extrema refer to the highest and lowest values a function attains on a given interval. To find these, evaluate the function at critical points and endpoints of the interval. The largest value is the absolute maximum, and the smallest is the absolute minimum.
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Finding Extrema Graphically

Critical Points

Critical points occur where the derivative of a function is zero or undefined. These points are potential locations for local extrema. For the function f(t) = 2 - |t|, consider where the derivative changes or is not defined, especially at t = 0 where the absolute value function has a cusp.
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Critical Points

Graphing Absolute Value Functions

Graphing absolute value functions involves understanding their V-shaped structure. For f(t) = 2 - |t|, the graph is an inverted V with a vertex at t = 0. Evaluate the function at endpoints t = -1 and t = 3, and at the vertex to determine the absolute extrema and their coordinates.
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Related Practice
Textbook Question

Identifying Extrema


In Exercises 61 and 62, the graph of f' is given. Assume that f is continuous, and determine the x-values corresponding to local minima and local maxima.


Textbook Question

Identifying Extrema


In Exercises 61 and 62, the graph of f' is given. Assume that f is continuous, and determine the x-values corresponding to local minima and local maxima.


Textbook Question

Absolute Extrema on Finite Closed Intervals


In Exercises 21–36, 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) = (2/3)x − 5, −2 ≤ x ≤ 3

Textbook Question

117. Suppose that the second derivative of the function y = f(x) isy" =(x+1)(x-2).

For what x-values does the graph of f have an inflection point?

Textbook Question

Finding Critical Points


In Exercises 41–50, determine all critical points and all domain endpoints for each function.


y = x² − 32√x

Textbook Question

Each of Exercises 67–88 gives the first derivative of a continuous function y=f(x). Find y'' and then use Steps 2–4 of the graphing procedure described in this section to sketch the general shape of the graph of f.

77. y' = cot(θ/2), for 0 < θ < 2π