Asked by doodle bob
Danika concludes that the following functions are inverses of each other because f(g(x)) = x. Do you agree with Danika? Explain your reasoning.
f(x) = |x|
g(x) = –x
All Answers 1
Answered by
GPT-5 mini
AI
No — Danika is wrong.
Two functions f and g are inverses only if both compositions are the identity: f(g(x)) = x for every x in the domain of g, and g(f(x)) = x for every x in the domain of f.
Here f(x) = |x| and g(x) = −x. Compute
- f(g(x)) = |−x| = |x|, which equals x only when x ≥ 0 (not for all x).
- g(f(x)) = −|x|, which equals x only when x ≤ 0 and x = 0 (in fact it fails for positive x).
Example: x = −1 gives f(g(−1)) = |1| = 1 ≠ −1, so f(g(x)) ≠ x in general.
Also f is not one-to-one (f(1) = f(−1)), so it does not have a two-sided inverse on all of R. (g happens to be its own inverse, but that doesn't make g the inverse of f.)
Two functions f and g are inverses only if both compositions are the identity: f(g(x)) = x for every x in the domain of g, and g(f(x)) = x for every x in the domain of f.
Here f(x) = |x| and g(x) = −x. Compute
- f(g(x)) = |−x| = |x|, which equals x only when x ≥ 0 (not for all x).
- g(f(x)) = −|x|, which equals x only when x ≤ 0 and x = 0 (in fact it fails for positive x).
Example: x = −1 gives f(g(−1)) = |1| = 1 ≠ −1, so f(g(x)) ≠ x in general.
Also f is not one-to-one (f(1) = f(−1)), so it does not have a two-sided inverse on all of R. (g happens to be its own inverse, but that doesn't make g the inverse of f.)
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