Question

Please, help
Prove \(f(x) =\dfrac{x}{(\sqrt (x^2+1)}\) is onto from R to (-1,1)

Answer 1

The previous part is f(x) is one to one and I got it. Just this part I need to confirm

Answer 2

what u got so far ?

Answer 3

So far, I get :
let y in (-1,1) , then \(\dfrac{x}{\sqrt{x^2+1}=y\)

Answer 4

if and only if \(x^2= y^2(x^2+1)\)
--> \(x^2 -y^2x^2 -y^2 =0\)

Answer 5

--> \(x^2(1-y^2) -y^2=0\)
so that \(x= \pm \dfrac{y}{\sqrt{1-y^2}}\)

Answer 6

and ... stuck

Answer 7

can we determine a min and max by chance? not that its proof, but could be usefl

Answer 8

take the limit as x to +- infinity

Answer 9

we can observe that there is no real value that will mess up the function; ie 0 on the bottom, or negative sqrt

Answer 10

if the slope if the function is always postive, and limits to 0 at the ends ...

Answer 11

well, always postive or always negative ... monotonic is the term maybe?

Answer 12

so??

Answer 13

For onto, you need to prove y takes all the values between (-1,1)

Answer 14

\[\lim_{x\to -inf}\frac{x}{\sqrt{x^2+1}}=-1\]
\[\lim_{x\to +inf}\frac{x}{\sqrt{x^2+1}}=1\]
\[f'=\frac{1}{(x^2+1)^{3/2}}\]
f' is never 0, and always positive

Answer 15

derivative test looks neat than finding x and bounding y

Answer 16

@amistre64 that is the part I prove f is one to one

Answer 17

if you go that route, try contradiction

Answer 18

it proves onto to me, since there is not x in R that is restricted, and that y = -1, y=0, y=1 are withing the range limits

Answer 19

that proves onto also because the lower and upper limits are proven to be -1 and +1

Answer 20

two birds in one shot !

Answer 21

f'(+- inf) to 0 as well

Answer 22

might not be book correct, but i like it :)

Answer 23

Please show me how \[\lim_{x\to -inf}\frac{x}{\sqrt{x^2+1}}=-1\]

Answer 24

\(\large x^2+1 \approx x^2\) as \(\large x\to \infty \)

Answer 25

\[\lim_{x\to -inf}\frac{x}{\sqrt{x^2+1}}=\lim_{x\to -inf}\frac{x}{x\sqrt{1+1/x^2}}=\lim_{x\to -inf}\frac{1}{\sqrt{1+0}}\] how to get -1

Answer 26

change of variables

y = -x

Answer 27

\[\large \lim_{x\to -\infty}\frac{x}{\sqrt{x^2+1}}= \lim_{y\to \infty}\frac{-y}{\sqrt{y^2+1}}\]

Answer 28

Thank you.

Answer 29

Do we have to show f(x) is continuous on the interval before taking lim?

Answer 30

A function is onto when its image equals its range
so you need to show that :-
1_ domain and range from (-1,1)
2_ both of f(x) and f^-1(x) are continues

Answer 31

One more question: to have f is onto from R to (-1,1) , we have to show that for all y in (-1,1) , x is defined in R
but, when solving \( y = \dfrac{x}{\sqrt{x^2+1}}\) in the interval (-1,1) , it means
\(-1<\dfrac{x}{\sqrt{x^2+1}}<1\)
and I can't define x satisfy that expression

Answer 32

I see nothing, girl :)

Answer 33

Maybe you could say $$\frac{x}{\sqrt{x^2+1}}=y$$and solve for x? This yields: $$x=\sqrt{\frac{y^2}{1-y^2}}$$ which is only a real number if: $$-1<y<1.$$

Answer 34

Best answer!! clear and neat.

Answer 35

Thank you so much. It makes sense to me.

Answer 36

You're welcome :)