What is the inverse of ƒ (x) = (2x + 1)² for x ≥ − 1/1/? ƒ−¹(x) = − ¹⁄/ (√x + 1) - for x ≥ 1/1/20 X (1 + x^² ) ² − = (x) ₁_ƒ° for x > 0 루 (1 - x^² ) ² = (x) ₁_ƒ ° for x ≥ 0 ƒ−¹(x) = ½ / (√x − 1) for x ≥ 1/1/1

Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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What is the inverse of the following equation?
What is the inverse of
ƒ (x) = (2x + 1)² for
x ≥ − 1/1/1?
2 -
ƒ−¹(x) = − ½¹⁄² (√x + 1)
for x ≥ 1/1/20
X >
。ƒ˜¹(x) = − 1 ² ( √√x + 1)
for x ≥ 0
of ¹(x) = ²/² (√x - 1)
for x > 0
ƒ−¹(x) = ½¼/² (√x − 1)
for X >
2
Transcribed Image Text:What is the inverse of ƒ (x) = (2x + 1)² for x ≥ − 1/1/1? 2 - ƒ−¹(x) = − ½¹⁄² (√x + 1) for x ≥ 1/1/20 X > 。ƒ˜¹(x) = − 1 ² ( √√x + 1) for x ≥ 0 of ¹(x) = ²/² (√x - 1) for x > 0 ƒ−¹(x) = ½¼/² (√x − 1) for X > 2
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