Textbook Question
In Exercises 41–44:
a. Find f⁻¹(x).
41. f(x) = 2x + 3, a = −1
Ch. 7 - Transcendental Functions
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- 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 Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 7, Problem 7.1.42a
In Exercises 41–44:
a. Find f⁻¹(x).
42. f(x) = (x + 2) / (1 − x), a = 1/2
Verified step by step guidance1
Start by writing the function as an equation with y: \(y = \frac{x + 2}{1 - x}\).
To find the inverse function \(f^{-1}(x)\), swap the roles of \(x\) and \(y\): \(x = \frac{y + 2}{1 - y}\).
Multiply both sides of the equation by the denominator \((1 - y)\) to eliminate the fraction: \(x(1 - y) = y + 2\).
Distribute \(x\) on the left side: \(x - xy = y + 2\).
Group all terms involving \(y\) on one side and constants on the other: \(x - 2 = y + xy\). Then factor \(y\) out: \(x - 2 = y(1 + x)\). Finally, solve for \(y\): \(y = \frac{x - 2}{1 + x}\), which is \(f^{-1}(x)\).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Inverse Functions
An inverse function reverses the effect of the original function, swapping inputs and outputs. To find f⁻¹(x), solve the equation y = f(x) for x in terms of y, then interchange x and y. The inverse exists only if the function is one-to-one (bijective) on the domain considered.
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Inverse Cosine
Solving Rational Equations
Finding the inverse of a rational function often requires solving equations involving fractions. This involves clearing denominators, isolating variables, and simplifying expressions carefully to avoid extraneous solutions or domain issues.
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5:02Solving Logarithmic Equations
Domain and Range Considerations
When finding inverses, it is crucial to consider the domain of the original function and the range of the inverse. Restrictions on x (like denominators not being zero) affect the domain, and the inverse’s domain corresponds to the original function’s range.
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Related Practice
Textbook Question
Evaluate the integrals in Exercises 67–74 in terms of
a. inverse hyperbolic functions.
67. ∫(from 0 to 2√3)dx/√(4+x²)
Textbook Question
Use reference triangles in an appropriate quadrant to find the angles in Exercises 1–8.
7. a. sec^(-1)(-√2)
Textbook Question
78. Which one is correct, and which one is wrong? Give reasons for your answers.
a. lim (x → 0) (x² - 2x) / (x² - sin x) = lim (x → 0) (2x - 2) / (2x - cos x) = lim (x → 0) 2 / (2 + sin x) = 2 / (2 + 0) = 1
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Textbook Question
Find the volumes of the solids in Exercises 135 and 136.
135. The solid lies between planes perpendicular to the x-axis at x=-1 and x=1. The cross-sections perpendicular to the x-axis are
a. circles whose diameters stretch from the curve y=-1/√(1+x²) to the curve y=1/√(1+x²).
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Textbook Question
[Technology Exercise] In Exercises 139–141, find the domain and range of each composite function. Then graph the compositions on separate screens. Do the graphs make sense in each case? Give reasons for your answers. Comment on any differences you see.
139. a. y=arctan(tan x)