Textbook Question
Roots (Zeros)
Show that the functions in Exercises 19–26 have exactly one zero in the given interval.
g(t) = √t + √(1 + t) − 4, (0, ∞)
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4. Applications of Derivatives
Differentials
3:38 minutesProblem 29a
Textbook Question
Finding Functions from Derivatives
Suppose that f'(x) = 2x for all x. Find f(2) if
a. f(0) = 0
Verified step by step guidance1
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.
Verified video answer for a similar problem:This video solution was recommended by our tutors as helpful for the problem above
Video duration:
3mKey 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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05:03Initial 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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4:26Evaluating Composed Functions
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