we have f: X→Y is a 1-1 function and Y is countable. Since f is 1-1, is it correct that this implies X ~ f(X) [which is f:X →f(X)] which further implies bijecton? Also, does this imply f(X) ⊆ Y?

Q: Are the following functions quasi-concave? Assume x ≥ 0 a) f(x) = a b) f(x) = a + cx², c < 0

A: Both questions are different so I soved 1st queation if you want 2nd one then repost it again.

Q: - h)? + k: a). f(x) = 2x² –12x+11; b) y = -r+x+:

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Q: x+7, -4 h er. f(0) = 7. f(1) = 8, f(2) = -5, and f(3) = -10, of b

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A: Given: f(x) = 5x2 + 3x – 26  is  θ(x 2).

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A:

Q: In each of the following, determine whether

A: We have to decide whether y is a function of x or not.

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A: We can find the domain of the given functions as below.