Are these functions injective/surjective etc?

**misqa88** · October 25th 2012, 10:34 AM

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....

**johnsomeone** · October 25th 2012, 12:13 PM

For question #1:
(c) is wrong (f(2) = ? f(-1) = ?)
Answer (c) before answering (a).
Answer (b) before answering (d).

he surjectivity of functions from $\mathbb{R}$ to $\mathbb{R}$ 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 $\mathbb{R},$ 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.

**Plato** · October 25th 2012, 12:20 PM

Originally Posted by 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 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....

This is a double post! DO NOT DO THAT!

Moreover this is not advanced algebra.

**misqa88** · October 25th 2012, 12:24 PM

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** · October 25th 2012, 12:32 PM

Originally Posted by misqa88

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.

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

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

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

**HallsofIvy** · October 25th 2012, 12:34 PM

Originally Posted by misqa88

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

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** · October 25th 2012, 12:53 PM

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** · October 25th 2012, 01:44 PM

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

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

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

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

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

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

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

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

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

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

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

**misqa88** · October 25th 2012, 01:54 PM

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

**johnsomeone** · October 25th 2012, 01:58 PM

Originally Posted by misqa88

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.

**misqa88** · October 25th 2012, 02:10 PM

I would say the function is injective because f(x; y) = (x + y; x) so the f(x1,x2) = (3,0.25)... don't really know

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