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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 4.2.28
Chapter 4, Problem 4.2.28

Finding Functions from Derivatives


Suppose that f(0) = 5 and that f'(x) = 2 for all x. Must f(x) = 2x + 5 for all x? Give reasons for your answer.

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To determine if f(x) = 2x + 5 for all x, we need to consider the information given: f'(x) = 2 for all x. This means that the derivative of f(x) is constant, indicating that f(x) is a linear function.
A linear function with a constant derivative of 2 can be expressed in the form f(x) = 2x + C, where C is a constant.
We are also given that f(0) = 5. We can use this initial condition to find the value of the constant C.
Substitute x = 0 into the function f(x) = 2x + C to find C: f(0) = 2(0) + C = 5, which simplifies to C = 5.
Thus, the function f(x) = 2x + 5 satisfies both the derivative condition f'(x) = 2 and the initial condition f(0) = 5. Therefore, f(x) = 2x + 5 for all x.

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Key Concepts

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

Derivative

The derivative of a function at a point measures the rate at which the function value changes as its input changes. It is the slope of the tangent line to the function's graph at that point. In this context, f'(x) = 2 indicates that the function has a constant rate of change, meaning it is linear.
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Derivatives

Initial Condition

An initial condition provides a specific value of the function at a particular point, which helps determine the constant of integration when finding the original function from its derivative. Here, f(0) = 5 is the initial condition that allows us to solve for the constant term in the linear function f(x).
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Initial Value Problems

Linear Function

A linear function is of the form f(x) = mx + b, where m is the slope and b is the y-intercept. Given f'(x) = 2, the function has a slope of 2. Using the initial condition f(0) = 5, we find the y-intercept b = 5, confirming that f(x) = 2x + 5 is indeed the function satisfying both the derivative and initial condition.
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Related Practice
Textbook Question

Finding Indefinite Integrals


In Exercises 17–56, find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation.


∫(x + 1) dx

Textbook Question

Finding Extrema from Graphs


In Exercises 11–14, match the table with a graph.


Textbook Question

Finding Indefinite Integrals


In Exercises 17–56, find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation.


∫7sin(θ/3) dθ

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.

y = 1 / (x² - 1)

Textbook Question

Use the linear approximation (1 + x)ᵏ ≈ 1 + kx to find an approximation for the function f(x) for values of x near zero.


c. f(x) = 1/√(1 + x)

1
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Textbook Question

In Exercises 53 and 54, find both dy/dx (treating y as a differentiable function of x) and dx/dy (treating x as a differentiable function of y). How do dy/dx and dx/dy seem to be related?


53. xy³ + x²y = 6