Question

find the range of the function \(f\)

Answer 1

let
\(\large \color{black}{\begin{align} f=\left\{\left(x,\ \dfrac{x^2}{1+x^2}\right):\ x\in \mathbb{R}\right\} \hspace{.33em}\\~\\
\end{align}}\)

be a function from \(\mathbb{R}\) to \(\mathbb{R}\) .Determine the range of \(f\)

Answer 2

seems to me all real numbers?

Answer 3

oh no

Answer 4

a fraction is only 1 if the numerator is equal to the denominator

Answer 5

no it cxannot have negative values

Answer 6

and it is unlikely that \(x^2+1=x^2\)

Answer 7

oh miss reading the function

Answer 8

also you notice that everything in sight is positive right?
there is an actual way to do this

Answer 9

or just wing it
the least it can be is zero, since everything is positive unless x is zero
and it cannot reach 1 since the numerator is smaller than the denominator

Answer 10

making the range
\[[0,1)\]
otherwise you have to set
\[y=\frac{x^2}{1+x^2}\] and solve for \(x\) which is a colossal pain

Answer 11

i was looking for a way which would be different from just observation

Answer 12

there is nothing wrong with thinking, not matter what you math teacher tells you

Answer 13

but yes, there is a way
solve
\[y=\frac{x^2}{1+x^2}\] for \(x\)

Answer 14

you will get a quadratic equation in \(y\)
you then take the discriminant and make sure it is greater than or equal to zero
enjoy, it is a pain in the neck

Answer 15

i mean \(y(y-1)\geq 0\)

Answer 16

is that correct

Answer 17

is it not y(1-y)>=0

Answer 18

looks right

Answer 19

ya this one ^

Answer 20

i think if you solve for \(x\)you get
\[x=\sqrt{\frac{y}{1-y}}\]

Answer 21

domain of that one is
\[[0,1)\] so range of your function is the same

Answer 22

really i think this is a butt load more work, than necessary
common sense often works in math too

Answer 23

but why this wont work \(y(1-y)>=0 \)

Answer 24

why strictly greater than?

Answer 25

oh i see you have \(\geq\)

Answer 26

it will be \(1\geq y\geq 0\)

Answer 27

but that is not right, because \(\geq\) includes 1

Answer 28

and there is no way for your function to be 1

Answer 29

thats why was asking why that method backstabbed

Answer 30

whereas solving
\[\frac{y}{1-y}\geq 0\] does not include 1

Answer 31

ok thnx

Answer 32

ok i see my mistake i mislead you

Answer 33

you do not get a quadratic, you solve
\[y=\frac{x^2}{1+x^2}\]
\[yx^2+y=x^2\\
yx^2-x^2=-y\\
(y-1)x^2=-y\\
x^2=\frac{y}{1-y}\\
x=\sqrt{\frac{y}{1-y}}\]

Answer 34

then solve \[\frac{y}{1-y}\geq0\] you get
\[[0,1)\]

Answer 35

sorry about that
i could never win against a king's gambit either, not matter whether i accepted or declined

Answer 36

haha

Answer 37

i have good record against kings gambit by black