Question

The function f is defined by

Answer 1

\[f:x \rightarrow \frac{ x }{ x^2+1 },x \in\mathbb{R} \]

Answer 2

If\[a\in\mathbb{R} \]and\[a \neq0\],find the image of \[\frac{ 1 }{ a }\]under f.

Answer 3

Deduce that f is not one-to-one.
Show that if \[a,b\in\mathbb{R} \] with\[a>b\ge\mathbb{R} \],then f(b)>f(a).

Answer 4

Deduce that,if the domain of f is restricted to the subset of\[\mathbb{R} \]given by\[{x:x\ge~1} \],then f is one-to-one.
State the range of f in this case.

Answer 5

\[\{x:x\ge~1\} \]

Answer 6

I got the answer for f(1/a)=a/(1+a^2)

Answer 7

Oh you simplified it, yayyy :)

Answer 8

LOL why am I getting all these medals? 0_o
I didn't do anything!!

Answer 9

just to lift up ur spirit... xD

Answer 10

Do u hv any ideas? @zepdrix

Answer 11

Showing One-to-One?
Mmm thinking :d

Answer 12

Grr math is hard :(
Still thinking.

And for some reason still getting medals t.t
Boy this is gonna be a big disappointment LOL

Answer 13

Ok let's try this maybe...
We'll start with this,
\(\large\rm a>b\)
and try to build our f(b) and f(a) in the inequality somehow.

Answer 14

Squaring each side,
\(\large\rm a^2>b^2\)

Adding 1 to each side,
\(\large\rm a^2+1>b^2+1\)
From here we can do some division.
Notice that since we have squares, we'll never be dividing by a negative.
So we don't have to worry about the inequality flipping.
\[\large\rm \frac{a^2+1}{b^2+1}>1\]And then divide by the a stuff,\[\large\rm \frac{1}{b^2+1}>\frac{1}{a^2+1}\]Then let's multiply by b,\[\large\rm \frac{b}{b^2+1}>\frac{b}{a^2+1}\]Ok so we've been able to show this,\[\large\rm f(b)>\frac{b}{a^2+1}\]

We might have some issues if one of the values, a OR b is negative.
I'm being a little sloppy I suppose.
I was a little unclear about what a > b > R means.

Answer 15

As a side note,
notice that \(\large\rm \dfrac{b}{a^2+1}>\dfrac{a}{a^2+1}\) because the numerator is larger (b is larger than a).

No no no I did that backwards... hmm hold on...
a is supposed to be larger than b.

Answer 16

Oh wait wait wait...
isn't that exactly what we want though?
Because this shows that the one-to-oneness is not followed!

Answer 17

Hmmm thinking...

Answer 18

Bahhh I dunno >.<
Ok I'm giving up for now.

Answer 19

Have you done the first part yet>

find the image of 1/a under f.

If so, what did you get?

Answer 20

i got f(1/a)=a/(1+a^2)

Answer 21

Okay.

Answer 22

The function y = x /( x² + 1 ) is not one-to-one because it fails the horizontal line test.

Look at the graph.

Answer 23

In the 4th post from the top, I do not know what it means to say that a > b ≥ ℝ
. How can a and b be larger than the set of real numbers? Is there a typo there?

Answer 24

f with restricted domain of x > 1

With restriction on domain, f is a one-to-one function.

Answer 25

The range may be 0 < y < 1/2.

I think that you are asked to give a formal argument such as the one begun by @zepdrix .

Correct this first: a > b ≥ ℝ

I think it is supposed to be: a > b where a and b are elements of ℝ.

Answer 26

ops...it is a typo @Directrix
it should be\[a>b \ge1\]

Answer 27

i got the range for function f...
yep,it is\[0<y \le \frac{ 1 }{ 2 }\]

Answer 28

Thank you @Directrix @zepdrix
^_^