Ch. 4 - Applications of Derivatives

Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo

Hass 15th Edition

Ch. 4 - Applications of Derivatives

Problema 4.1.35

Capitolo 4, Problema 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

Guida verificata passo dopo passo

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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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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Percorso guidato

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Average Value of a Function

Pratica correlata

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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.

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Domanda del libro di testo

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.

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Finding Indefinite Integrals

Find the indefinite integrals (most general antiderivatives) in Exercises 73–88. You may need to try a solution and then adjust your guess. Check your answers by differentiation.

∫ 𝓍³ (1 + 𝓍⁴ )⁻¹/⁴ d𝓍

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Domanda del libro di testo

Identifying Extrema

In Exercises 15–18:

a. Find the open intervals on which the function is increasing and those on which it is decreasing.

b. Identify the function’s local and absolute extreme values, if any, saying where they occur.

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Initial Value Problems

Find the curve y = f(x) in the xy-plane that passes through the point (9,4) and whose slope at each point is 3√x.

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Finding Indefinite Integrals

Find the indefinite integrals (most general antiderivatives) in Exercises 73–88. You may need to try a solution and then adjust your guess. Check your answers by differentiation.

∫ sec θ/3 tan θ/3 dθ

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