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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077당신이 사용하는 게 아니라요?교과서 변경
7장, 문제 7.1.42a

In Exercises 41–44:
a. Find f⁻¹(x).


42. f(x) = (x + 2) / (1 − x), a = 1/2

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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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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

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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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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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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