Question

we have a continues function at [0,1] , f(0) = 0 and f(1)=1 , how do i prove that there is a number c from ]0,1[ achieve f(c) = (1-c)/(1+c) ?

Answer 1

What do you think @Zarkon i know that we should use the intermediate values theorem but how ?

Answer 2

look at \[g(x)=f(x)-\frac{1-x}{1+x}\]

Answer 3

Great then ?

Answer 4

what is g(0) and g(1)?

Answer 5

g(0) = f(0) - 1 ?

Answer 6

yes

=-1

Answer 7

g(1) = f(1)

Answer 8

=1

Answer 9

exactly

Answer 10

you have it from here?

Answer 11

then ?

Answer 12

use the IVT

Answer 13

for g or f ?

Answer 14

g

Answer 15

g(0)<0 and g(1)>0 so...

Answer 16

g(0)*g(1) < 0

Answer 17

so there is a c for g , what about f ?

Answer 18

got there is a c such that g(c)=0


so \[f(c)-\frac{1-c}{1+c}=0\]

Answer 19

mmmm cool

Answer 20

@Zarkon how did you think about the g(x) function , is there a princible ?