Textbook Question
Finding Functions from Derivatives
In Exercises 31–36, find all possible functions with the given derivative.
a. y′ = x
Ch. 4 - Applications of Derivatives
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 4, Problem 29b
Finding Functions from Derivatives
Suppose that f'(x) = 2x for all x. Find f(2) if
b. f(1) = 0
Verified step by step guidance1
To find the function f(x) from its derivative f'(x) = 2x, we need to integrate the derivative. The integral of f'(x) = 2x with respect to x is f(x) = ∫2x dx.
Perform the integration: ∫2x dx = x^2 + C, where C is the constant of integration. This gives us the general form of the function: f(x) = x^2 + C.
We are given the initial condition f(1) = 0. Use this information to find the constant C. Substitute x = 1 into the function: f(1) = 1^2 + C = 0.
Solve for C: 1 + C = 0, which implies C = -1.
Now that we have determined C, the specific function is f(x) = x^2 - 1. To find f(2), substitute x = 2 into this function: f(2) = 2^2 - 1.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Antiderivatives
An antiderivative of a function is another function whose derivative is the original function. In this context, finding the antiderivative of f'(x) = 2x involves determining a function f(x) such that its derivative is 2x. The general form of the antiderivative of 2x is x^2 + C, where C is a constant.
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Antiderivatives
Initial Conditions
Initial conditions are specific values given for a function at a particular point, which help determine the constant of integration when finding an antiderivative. Here, the initial condition f(1) = 0 allows us to solve for the constant C in the antiderivative f(x) = x^2 + C, ensuring the function satisfies this condition.
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Definite Evaluation
Definite evaluation involves using the determined function, including the constant of integration, to find the value of the function at a specific point. After finding the antiderivative and constant, we use f(x) = x^2 - 1 to evaluate f(2), which involves substituting x = 2 into the function to find the specific value of f(2).
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05:43Definition of the Definite Integral
Related Practice
Textbook Question
Calculate the first derivatives of ƒ(𝓍) = 𝓍²/ (𝓍² + 1) and g(𝓍) = ―1/ (𝓍² + 1) . What can you conclude about the graphs of these functions?
Textbook Question
Finding Functions from Derivatives
Suppose that f'(x) = 2x for all x. Find f(2) if
a. f(0) = 0
Textbook Question
Finding Functions from Derivatives
In Exercises 31–36, find all possible functions with the given derivative.
b. y′ = x²
Textbook Question
In Exercises 9–66, graph the function using appropriate methods from the graphing procedures presented just before Example 9, identifying the coordinates of any local extreme points and inflection points. Then find coordinates of absolute extreme points, if any.
30. y = (x² - 4) / (x² - 2)
Textbook Question
Finding Functions from Derivatives
Suppose that f(−1) = 3 and that f'(x) = 0 for all x. Must f(x) = 3 for all x? Give reasons for your answer.