1. Let f : R -> R be the function given by f(x) = x^3 - x^2 - 2x.
(a) The function f is injective.
(b) The function f is surjective.
(c) f^-1({0}) = {0}
(d) -1 ∈ f(R)
2. Let f : R2 -> R2 be a function given by f(x1; x2) = (3; x1 - 2*x2)
(a) The function f is injective
(b) The function f is surjective
(c) The image of f is the line dened by y1 - 2*y2 = 3
(d) f^-1({(y1; y2) ∈ R2 , y2 = 0}) is the line dened by x1 - 2*x2 = 0.
Please help! I don't get it at all....
1) yes it is one-to-one (so the function is injective)
2) no the graph doesn't pass the horizontal line test
3) do you mean the pint (0,0)? it's above the graph
4) yes the line cuts the graph
Oh sorry! the function is not injective because the y-values are not unique for x-values, but it is subjective because it passes the horizontal line test. Is it always the case? Can I always check the surjectivity of a function like that? Sorry for all those questions but you are the most helpful so far.
1. Let f : R -> R be the function given by f(x) = x^3 - x^2 - 2x.
(a) The function f is injective.
(b) The function f is surjective.
(c) f^-1({0}) = {0}
(d) -1 ∈ f(R)
Guys, for surjective or not, find maxima and minima, then see if those are bounded or not (not +infinity and -infinity). If those are bounded, then it is not surjective, otherwise it is surjective.
Salahuddin
Maths online
That is correct. But in our examples, those are continuous and differentiable, etc... right. No harm of using it as long as those are polynomials, and such.
Salahuddin
Maths online