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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

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.

69. y' = x(x - 3)²

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Textbook Question

Initial Value Problems


Solve the initial value problems in Exercises 71–90.


dy/dx = 2x − 7, y(2) = 0

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Textbook Question

Finding Position from Velocity or Acceleration


Exercises 45–48 give the acceleration a=d²s/dt², initial velocity, and initial position of an object moving on a coordinate line. Find the object’s position at time t.


a = 9.8, v(0) = −3, s(0) = 0

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Textbook Question

Theory and Examples


Determine the values of constants a and b so that f(x) = ax² + bx has an absolute maximum at the point (1,2).

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Textbook Question

30. Find a positive number for which the sum of its reciprocal and four times its square is the smallest possible.

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Textbook Question

Theory and Examples

67. An inequality for positive integers Show that if a, b, c, and d are positive integers, then

[(a^2+1)(b^2+1)(c^2+1)(d^2+1)]/abcd ≥ 16

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