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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 7, Problem 7.1.41a

In Exercises 41–44:
a. Find f⁻¹(x).


41. f(x) = 2x + 3, a = −1

Verified step by step guidance
1
Start with the given function: \(f(x) = 2x + 3\).
To find the inverse function \(f^{-1}(x)\), replace \(f(x)\) with \(y\): \(y = 2x + 3\).
Swap the variables \(x\) and \(y\) to get \(x = 2y + 3\); this step reflects the idea that the inverse function reverses the roles of inputs and outputs.
Solve the equation \(x = 2y + 3\) for \(y\): subtract 3 from both sides to get \(x - 3 = 2y\), then divide both sides by 2 to isolate \(y\): \(y = \frac{x - 3}{2}\).
Rewrite \(y\) as \(f^{-1}(x)\) to express the inverse function: \(f^{-1}(x) = \frac{x - 3}{2}\).

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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. For a function f(x), its inverse f⁻¹(x) satisfies f(f⁻¹(x)) = x. Finding the inverse involves solving the equation y = f(x) for x in terms of y.
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Solving Linear Equations

To find the inverse of a linear function like f(x) = 2x + 3, you solve for x by isolating it on one side of the equation. This involves algebraic manipulation such as subtracting constants and dividing by coefficients to express x as a function of y.
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Domain and Range Considerations

When finding an inverse, it's important to consider the domain and range of the original function and its inverse. For linear functions, the domain and range are typically all real numbers, but specifying points like a = -1 helps evaluate the inverse at particular values.
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