# Are these functions injective/surjective etc?

#### misqa88

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 de ned by y1 - 2*y2 = 3
(d) f^-1({(y1; y2) ∈ R2 , y2 = 0}) is the line de ned by x1 - 2*x2 = 0.

Please help! I don't get it at all.... #### johnsomeone

For question #1:
(c) is wrong (f(2) = ? f(-1) = ?)

he surjectivity of functions from $$\displaystyle \mathbb{R}$$ to $$\displaystyle \mathbb{R}$$ is asking if it ever goes to minus-infinity, and if it ever goes to infinity. That's including when x goes to plus or minus infinity. Basically, you're asking if it's bounded above, and if it's bounded below.

Of course, there's a little bit more involved. For instance, the function y = 1/x is continuous everywhere except x=0, and is neither bounded above nor bounded below, but isn't surjective onto $$\displaystyle \mathbb{R},$$ because it never takes the value 0.

Technically, you'll use the Intermediate Value Theorem and an analysis of the function's continuity to derive that it's surjective (if it is surjective).

For question 2, it's simple enough to directly check if f is injective, and chekc if f is surjective. However, if you know linear algebra, a useful observation is that L(x1, x2) = f(x1, x2) - (3, 0) defines a linear map L. So it's a simple argument to see that whether f is injective (or surjective) will be the same as whether L is injective (or surjective) - and with linear maps, there are some tools you can use to answer that question.

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#### Plato

MHF Helper
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 de ned by y1 - 2*y2 = 3
(d) f^-1({(y1; y2) ∈ R2 , y2 = 0}) is the line de ned by x1 - 2*x2 = 0.

Please help! I don't get it at all.... This is a double post! DO NOT DO THAT!

Moreover this is not advanced algebra.

#### misqa88

hi, thanks for replying. c) is what I got from my lecturer so it must be correct. Could please explain it in more details? I know the theory but I dont know how to use it in practice as my lecturer is useless.
Thx.

#### johnsomeone

c) is what I got from my lecturer so it must be correct.
Either you mistyped the problem or your lecturer made a mistake. We're all human - we all make mistakes. That includes me, you, your lecturer, everyone. I've seen a Fields Medalist make simple mistakes at the blackboard.

$$\displaystyle \text{If } f(x) = x^3 - x^2 - 2x, \text{ then } f(x) = x(x-2)(x+1),$$

$$\displaystyle \text{so } f^{-1}(\{0\}) = \{0, 2, -1 \}.$$

Also - please heed Plato's post. It's inconsiderate to double post, and you should try to put your posts in the correct category.

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#### HallsofIvy

MHF Helper
hi, thanks for replying. c) is what I got from my lecturer so it must be correct.
You have two problems, both having a "c". Which are you talking about?

If the first one, $$\displaystyle f(x)= x^3- x^2- 2x$$ then it shoud be easy to see that $$\displaystyle f(0)= 0^3- 0^2- 2(0)= 0$$, $$\displaystyle f(2)= 2^3- 2^2- 2(2)= 8- 4- 4= 0$$, $$\displaystyle f(-1)= (-1)^3- (-1)^2-2(-1)= -1- 1+ 2= 0$$.

Could please explain it in more details? I know the theory but I dont know how to use it in practice as my lecturer is useless.
Thx.
So you too have one of those lecturers who don't think their job is to teach you integer arithmetic?

#### misqa88

Sorry that's my first time here, please excuse my mistakes. HallsofIvy I was talking about c) for Q1. I had math 10 years ago at school and I don't really remember how to do this things. My lecturer is not good in explaining and I can't find any book explaining examples step by step so I can apply those steps in different questions. I was hoping someone can explain it to me in more detail here. For Q1 I draw a graph and it look like a polynomial graph... Please help.

#### johnsomeone

$$\displaystyle \text{1c) As my previous post demonstrates, } f^{-1}(\{ 0 \}) = \{ 0, 2, -1 \}.$$

$$\displaystyle \text{1a) From 1c, have that } f \text{ is not injective, since } f(0) = f(2).$$

$$\displaystyle \text{1b) } \lim_{x \to \infty } f(x) = \lim_{x \to \infty } x^3 - x^2 - 2x = \infty. \text{ Likewise } \lim_{x \to -\infty } f(x) = - \infty.$$

$$\displaystyle \text{Therefore f is surjective onto } \mathbb{R}, \text{ using the IVT and that } f \text{ is continuous on } \mathbb{R}.$$

$$\displaystyle \text{[[That's applying the Intermediate Value Theorem (IVT). Let } c \in \mathbb{R}.$$

$$\displaystyle \text{Then, using }\lim_{x \to -\infty } f(x) = -\infty \text{ and } \lim_{x \to \infty } f(x) = \infty,$$

$$\displaystyle \text{have that there exists some } x_0, x_1, \text{ such that } f(x_0) < c < f(x_1).$$

$$\displaystyle \text{Then, since } f \text{ is continuous between } x_0 \text{ and } x_1, \text{ have that there exists } x_2 \text{ between } x_0 \text{ and } x_1$$

$$\displaystyle \text{such that } f(x_2) = c. \text{ That's the IVT. Thus } f \text{ hits every value }c \in \mathbb{R}, \text{and so } f \text{ is surjective.]]}$$

$$\displaystyle \text{1d) From 1b, } f \text{ is surjective onto } \mathbb{R}.$$

$$\displaystyle \text{Thus there exists } a \in \mathbb{R} \text{ such that } f(a) = -1. \text{ Therefore } -1 \in f(\mathbb{R}).$$

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#### misqa88

Thank you so much! what happens with the example 2 when the function looks like that f(x1; x2) = (3; x1 - 2*x2)?

#### johnsomeone

what happens with the example 2 when the function looks like that f(x1; x2) = (3; x1 - 2*x2)?
You tell me.
What's the definition of injective? Can you show that f is injective - or find a counter-example showing that f is not surjective?
Likewise for f being surjective or not.