Domanda del libro di testo
Each of Exercises 25–36 gives a formula for a function y=f(x). In each case, find f^(-1)(x) and identify the domain and range of f^(-1). As a check, show that f(f^(-1)(x))=f^(-1)(f(x))=x.
f(x) = x² − 2x, x ≤ 1
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Ch. 7 - Transcendental Functions
Hass15th EditionThomas' CalculusISBN: 9780137616077
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TOC.tableOfContents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions490
- Ch. 8 - Techniques of Integration399
- Ch. 9 - First-Order Differential Equations66
- Ch. 10 - Infinite Sequences and Series132
- Ch. 11 - Parametric Equations and Polar Coordinates58
Capitolo 7, Problema 7.6.117
Solve the initial value problems in Exercises 115–120.
117. dy/dx = 1/(x√(x² - 1)), x > 1; y(2) = π
Guida verificata passo dopo passo1
Identify the given differential equation: \(\frac{dy}{dx} = \frac{1}{x \sqrt{x^{2} - 1}}\) with the initial condition \(y(2) = \pi\) and domain \(x > 1\).
Rewrite the differential equation in differential form: \(dy = \frac{1}{x \sqrt{x^{2} - 1}} \, dx\).
Integrate both sides with respect to \(x\): \(y = \int \frac{1}{x \sqrt{x^{2} - 1}} \, dx + C\), where \(C\) is the constant of integration.
To solve the integral \(\int \frac{1}{x \sqrt{x^{2} - 1}} \, dx\), consider using a trigonometric substitution such as \(x = \sec \theta\), which simplifies the square root expression.
After finding the antiderivative, apply the initial condition \(y(2) = \pi\) to solve for the constant \(C\), then write the explicit solution for \(y\).
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Separable Differential Equations
A separable differential equation can be written as dy/dx = g(x)h(y), allowing the variables y and x to be separated on opposite sides of the equation. This enables integration with respect to each variable independently, facilitating the solution of the differential equation.
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06:06
Solving Separable Differential Equations
Integration Techniques Involving Inverse Trigonometric Functions
Integrals involving expressions like 1/(x√(x² - 1)) often lead to inverse trigonometric functions such as arcsec or arccos. Recognizing these forms and applying appropriate substitution or standard integral formulas is essential to solve the integral correctly.
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04:51Integrals Resulting in Inverse Trig Functions
Initial Value Problems (IVP)
An initial value problem specifies a differential equation along with a condition y(x₀) = y₀. Solving an IVP involves finding the general solution and then using the initial condition to determine the particular constant, yielding a unique solution curve.
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Percorso guidato
05:03Initial Value Problems
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Evaluate the integrals in Exercises 77–90.
79. ∫(from -1 to 0)6dt/√(3-2t-t²)
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Domanda del libro di testo
Use l’Hôpital’s rule to find the limits in Exercises 7–52.
24. lim (x → π/2) (ln(csc x)) / (x - (π/2))²
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In Exercises 5–8, show that each function is a solution of the given initial value problem.
5. Differential Equation: 2y + y' = 4x + 2
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Domanda del libro di testo
Theory and Applications
L’Hôpital’s Rule does not help with the limits in Exercises 69–76.
Try it—you just keep on cycling. Find the limits some other way.
73. lim (x → ∞) (2^x - 3^x) / (3^x + 4^x)
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Domanda del libro di testo
In Exercises 7–26, find the derivative of y with respect to x, t, or θ, as appropriate.
y = e^(5-7x)
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