Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
An injective function?
Need to prove that x^3 + 3x is an injective function ( 1-to-1 and onto) algebraically (no calculus)
2 Answers
- PopeLv 73 years agoFavorite Answer
Let f(x) = x³ + 3x.
Consider real numbers a and b, where |a| ≤ |b|.
Suppose f(a) = f(b).
f(a) = f(b)
a³ + 3a = b³ + 3b
a³ - b³ + 3a - 3b = 0
(a - b)(a² + ab + b²) + 3(a - b) = 0
(a - b)(a² + ab + b² + 3) = 0
a - b = 0 or a² + ab + b² + 3 = 0
|a| ≤ |b|
|ab| ≤ b² ... The area of a rectangle is no greater than the square of its longest side.
b² - |ab| ≥ 0
b² + ab ≥ 0
a² + ab + b² + 3 > 0
a² + ab + b² + 3 ≠ 0
a - b = 0
a = b
It is shown that f(a) = f(b) only if a = b. Therefore, the function f is injective.
As for being onto (surjective), no codomain was declared in the function definition. It would then have to default to its range, so naturally it is surjective.
