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Ch. 4 - Applications of Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoHass 15th EditionCh. 4 - Applications of DerivativesProblema 38
Capitolo 4, Problema 38

Finding Functions from Derivatives


In Exercises 37–40, find the function with the given derivative whose graph passes through the point P.


g'(x) = 1 / x² + 2x, P(−1, 1)

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To find the original function g(x) from its derivative g'(x), we need to integrate the derivative. Start by setting up the integral of g'(x): ∫(1/x² + 2x) dx.
Break down the integral into two separate integrals: ∫(1/x²) dx + ∫(2x) dx.
Integrate each term separately. The integral of 1/x² is -1/x, and the integral of 2x is x². So, the antiderivative is -1/x + x² + C, where C is the constant of integration.
Use the given point P(-1, 1) to find the constant C. Substitute x = -1 and g(x) = 1 into the antiderivative: 1 = -1/(-1) + (-1)² + C.
Solve the equation from the previous step to find the value of C. Substitute this value back into the antiderivative to get the final expression for g(x).

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Antiderivatives

Antiderivatives, or indefinite integrals, are functions that reverse the process of differentiation. To find a function from its derivative, you need to determine its antiderivative. This involves integrating the given derivative function, which in this case is g'(x) = 1/x² + 2x, to find g(x).
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Antiderivatives

Initial Conditions

Initial conditions are specific values that allow us to find the particular solution of an antiderivative. Given a point P(-1, 1), we use this to determine the constant of integration after finding the general antiderivative. This ensures the function passes through the specified point, making it unique.
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Initial Value Problems

Integration Techniques

Integration techniques are methods used to find antiderivatives. For g'(x) = 1/x² + 2x, you can integrate each term separately: the integral of 1/x² is -1/x, and the integral of 2x is x². Understanding these techniques is crucial for solving the problem and finding the correct function g(x).
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Introduction to Indefinite Integrals
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Finding Functions from Derivatives


In Exercises 37–40, find the function with the given derivative whose graph passes through the point P.


r'(t) = sec t tan t − 1, P(0, 0)

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Graphs and Graphing


Graph the curves in Exercises 33–42.


y = 𝓍³ (8―𝓍 )

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Finding Functions from Derivatives


In Exercises 37–40, find the function with the given derivative whose graph passes through the point P.


r'(θ) = 8 − csc²θ, P(π/4, 0)

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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.

36. y = (x³ + x - 2) / (x - x²)

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Finding Position from Velocity or Acceleration


Exercises 41–44 give the velocity v = ds/dt and initial position of an object moving along a coordinate line. Find the object’s position at time t.


v = 9.8t + 5, s(0) = 10

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Finding Functions from Derivatives


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


c. y' = sin (2t) + cos (t/2)

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