Question

Functions

Answer 1

\(\large \color{black}{\begin{align}& f:\mathbb{R}\rightarrow \mathbb{R} ,\ \{x,y\} \in \mathbb{R} \hspace{1.5em}\\~\\
& f(x+f(y))=f(x)+y-1, \ f(0)=1 , f(-1)= ?\hspace{1.5em}\\~\\

\end{align}}\)

Answer 2

if y is 0 then we have
\[f(x+f(0))=f(x)+0-1 \\ f(x+1)=f(x)-1 \text{ since } f(0)=1 \\ \text{ now enter in } x \text{ as -1 }\]

Answer 3

I'm thinking that since functions have only one output for any given input, and we are dealing with a function f that can take an argument x and y, that f is a bijection. Therefore, for every input there is exactly one output and for every output there is exactly one input.

So f(y) must equal to x:

$$
f(x+f(y))=f(x)+y-1\\
f(x+x)=f(x)+f(x)-1\\
f(2x)=2f(x)-1\\
f(x)=\cfrac{f(2x)+1}{2}
$$
f(0)=1
Let's check
$$
f(0)=1=\cfrac{f(2\cdot 0)+1}{2}=\cfrac{f(0)+1}{2}=1\\
$$
So
$$

f(-1)=\cfrac{f(-2)+1}{2}\\
f(-2)=\cfrac{f(-4)+1}{2}\\
...
$$

Not sure if this is the right path, but we can probably find a solution to the recursive equation...

Answer 4

@freckles path sounds more sane and much easier

Answer 5

wow , so quick

Answer 6

so we just solve this type of questions by putting random numeric value

Answer 7

ones that will help :p

Answer 8

like we know f(0)=1
so wanted to try to use that
and then we knew we wanted to find f(-1)
so I play with some 0's and -1's
or what would the insides of f( ) 0 or -1
but I played with f(0) first since I knew its output