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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 29a
Chapter 4, Problem 29a

Finding Functions from Derivatives


Suppose that f'(x) = 2x for all x. Find f(2) if


a. f(0) = 0

Verified step by step guidance
1
Start by recognizing that f'(x) = 2x is the derivative of the function f(x). To find f(x), we need to integrate f'(x).
Set up the integral of f'(x) = 2x with respect to x: ∫2x dx.
Perform the integration: The integral of 2x with respect to x is x^2 + C, where C is the constant of integration.
Use the initial condition f(0) = 0 to find the constant C. Substitute x = 0 into the equation f(x) = x^2 + C, giving f(0) = 0^2 + C = 0, which implies C = 0.
Now that we have f(x) = x^2, substitute x = 2 into the function to find f(2): f(2) = 2^2.

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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 f'(x) is a function f(x) whose derivative is f'(x). To find f(x) from f'(x) = 2x, we integrate 2x with respect to x, resulting in f(x) = x^2 + C, where C is the constant of integration.
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Antiderivatives

Initial Conditions

Initial conditions are specific values that allow us to determine the constant of integration when finding an antiderivative. Given f(0) = 0, we substitute x = 0 into f(x) = x^2 + C, yielding 0 = 0^2 + C, which implies C = 0.
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Initial Value Problems

Evaluating Functions

Once the function f(x) is determined, we can evaluate it at specific points. With f(x) = x^2 and C = 0, we find f(2) by substituting x = 2, resulting in f(2) = 2^2 = 4.
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Evaluating Composed Functions
Related Practice
Textbook Question

As a result of a heavy rain, the volume of water in a reservoir increased by 1400 acre-ft in 24 hours. Show that at some instant during that period the reservoir’s volume was increasing at a rate in excess of 225,000 gal/min. (An acre-foot is 43,560 ft³, the volume that would cover 1 acre to the depth of 1 ft. A cubic foot holds 7.48 gal.) 

Textbook Question

Finding Functions from Derivatives


In Exercises 31–36, find all possible functions with the given derivative.


a. y′ = x

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

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.

Textbook Question

Finding Functions from Derivatives


Suppose that f'(x) = 2x for all x. Find f(2) if


b. f(1) = 0