Funky Function

One can do this the general way, or just plug in -4. I'll do both here. $f(-4)+f(\frac {(-4)-1}{(-4)})=\frac{1}{(-4)}$ or $f(-4)+f(\frac {5}{4})=-\frac{1}{4}$

$f(\frac {5}{4})+f(\frac {\frac {5}{4}-1}{\frac {5}{4}})=\frac{1}{\frac {5}{4}}$ or $f(\frac {5}{4})+f(\frac {1}{5})=\frac{4}{5}$

$f(\frac {1}{5})+f(\frac {\frac {1}{5}-1}{\frac {1}{5}})=\frac{1}{\frac {1}{5}}$ , or $f(\frac {1}{5})+f(-4)=5$

$f(-4)+f(\frac {5}{4})+f(\frac {1}{5})+f(-4)=5+(-\frac{1}{4})$

$2f(-4)+f(\frac{5}{4})+f(\frac{1}{5})=\frac{19}{4}$ But $f(\frac {5}{4})+f(\frac {1}{5})=\frac{4}{5}$ , so

$2f(-4)+\frac{4}{5}=\frac{19}{4}$

$f(-4)=\frac{79}{40}$ , and since $gcd(79,40)=1$ , they are coprime, and $\frac{a}{b}=\frac{79}{40} \Rightarrow a+b=79+40=119$

This is the general method, where we solve for f(x): $f(x)+f(\frac {x-1}{x})=\frac{1}{x}$ . Let this equation be $(A)$ . Then,

$f(\frac {x-1}{x})+f(\frac {\frac {x-1}{x}-1}{\frac {x-1}{x}})=\frac{1}{\frac{x-1}{x}}$ , or $f(\frac {x-1}{x})+f(\frac {1}{1-x})=\frac {x}{x-1}$ Let this equation be $(B)$ . The last iteration will be:

$f(\frac {1}{1-x})+f(\frac {\frac {1}{1-x}-1}{\frac {1}{1-x}})=\frac{1}{\frac {1}{1-x}}$ , or $f(\frac {1}{1-x})+f(x)=1-x$ Let this be equation $(C)$ . Now add equations $(A)$ and $(C)$ .

$f(x)+f(\frac {x-1}{x})+f(\frac {1}{1-x})+f(x)=\frac{1}{x}+1-x$

$2f(x)+f(\frac {x-1}{x})+f(\frac {1}{1-x})=\frac{x+1}{x}$ Now, substitute the two last terms on the right side using $(B)$ :

$2f(x)+\frac {x}{x-1}=\frac{x+1}{x}$ With a little algebra, we get

$f(x)=\frac{x^3-x^2+1}{2x(1-x)}$ , and

$f(-4)=\frac{(-4)^3-(-4)^2+1}{2(-4)(1-(-4))}=\frac{79}{40}$

Let $g(x)=\frac{x-1}{x}$ . Some scratch-work reveals that, for $x\neq0,1$ , $g(g(g(x)))=x$ . Keeping this in mind, and iteratively plugging in $x, g(x), g(g(x)),$ for $x$ , we get

$f(x)+f(g(x))=\frac{1}{x}$ ,
$f(g(x))+f(g(g(x)))=\frac{1}{g(x)}$ ,
$f(g(g(x)))+f(g(g(g(x))))=f(g(g(x)))+f(x)=\frac{1}{g(g(x))}$ ,

where we used the identity obtained for $g(x)$ in the last equation. Since we can calculate $\frac{1}{x}$ , $\frac{1}{g(x)}$ , and $\frac{1}{g(g(x))}$ , the three equations above give us linear system of equations for $f(x)$ , $f(g(x))$ , and $f(g(g(x)))$ . Adding the first and third equations and then subtracting the second gives us $2f(x)=\frac{1}{x}+\frac{1}{g(g(x))}-\frac{1}{g(x)} \Rightarrow f(x)=\frac{1}{2}(\frac{1}{x}+\frac{1}{g(g(x))}-\frac{1}{g(x)})$ . Plugging in $x=-4$ yields $f(-4)=\frac{79}{40}=\frac{a}{b}$ . $79=a$ and $40=b$ are coprime, giving us the answer $a+b=79+40=119$ .

We are given $f(x)+f(\frac{x-1}{x})=\frac1x$ . (1)

We substitute $\frac{x-1}{x}$ into $x$ in (1) to get:

$f(\frac{x-1}{x})+f\left(\frac{\frac{x-1}{x}-1}{\frac{x-1}{x}}\right)=\frac{1}{\frac{x-1}{x}}$ , or

$f(\frac{x-1}{x})+f(\frac{1}{1-x})=\frac{x}{x-1}$ . (2)

Now we substitute $\frac{1}{1-x}$ into $x$ in (1) to get:

$f(\frac{1}{1-x})+f\left(\frac{\frac{1}{1-x}-1}{\frac{1}{1-x}}\right)=\frac{1}{\frac{1}{1-x}}$ , or

$f(\frac{1}{1-x})+f(x)=1-x$ . (3)

By subtracting (2) from (1), we have:

$f(x)-f(\frac{1}{1-x})=\frac{x^2-x+1}{x(1-x)}$ . (4)

We then add (3) and (4) to have:

$2f(x)=\frac{x^2-x+1}{x(1-x)}+(1-x)$ , or

$f(x)=\frac{x^3-x^2+1}{2x(1-x)}$ .

Hence, $f(-4)=\frac{79}{40}$ , and $79+40=119$ .

*We can also solve for $f(-4)$ without using the system of 3 functional equations by substituting $-4$ in $x$ . We'll get $f(-4)+f(\frac54)=-\frac14$ . Then we substitute $\frac54$ in $x$ to get $f(\frac54)+f(\frac15)=\frac45$ . Finally we substitute $\frac15$ in $x$ to get $f(\frac15)+f(-4)=5$ and solve the system of equations for $f(-4)$ .

when we substitute values of x = -4 , 1/5 , 5/4, we get three equations with three unknowns and then we can solve easily for f(-4).

We remark that $f(x) + f\left(1- \frac{1}{x}\right) = \frac{1}{x}$ inspires us to consider $g(x) = 1- \frac{1}{x}$ , which is well-known to satisfy $g(g(g(x))) = x$ . While we admit that knowing this fact helps inspire the solution, it is not difficult to quickly to show this property. The original equation becomes $f(x) + f(g(x)) = \frac{1}{x}$ . Now, take $s,t \in \mathbb{R}$ such that $g(s) = x$ and $g(g(t)) = x$ . Making use of our original identity, $g(g(g(x))) = x$ , this function must also satisfy $f(g(x)) + f(g(g(x))) = \frac{1}{g(x)}$ and $f(g(g(x))) + f(x) = \frac{1}{g(g(x))}$ . We have three equations, from which we can isolate $f(x)$ as $f(x) = \frac{\frac{1}{x} - \frac{1}{g(x)} + \frac{1}{g(g(x))}}{2}$ . Then, we simply substitute $x=-4$ , and after some algebra, we arrive at $f(-4) = \frac{\frac{1}{-4} - \frac{1}{5/4} + \frac{1}{1/5}}{2} = \frac{79}{40}$ so $a+b = 119$ .

x = -4, so f(-4) + f(5/4) = -1/4, x = 5/4, f(5/4) + f(1/5) = 4/5, so f(-4) - f(1/5) = -21/20. x = 1/5, f(1/5) + f(-4) = 5, elimination of f(1/5),we get 2f(-4) = 79/20 So the answer f(-4) = 79/40, and a + b = 79 + 40 = 119

notice that f(1/5) + f(-4) = 5 - (1) f(5/4) + f(1/5) = 4/5 - (2) f(-4) + f(5/4) = -1/4 - (3) (1) - (2), we get f(-4) - f(5/4) = 21/5 - (4) (3) + (4), 2f(-4) = 79/20 f(-4) = 79/40 f (-4) = 79+40 = 119

when x = 1/5, f(1/5) + f(-4) = 5 when x= 5/4, f(5/4) + f(1/5) = 4/5. when x= -4, f(-4) + f(5/4) = -1/4, there are three unknowns and three equations, we can solve them to get, f(-4) = 79/40... therefore a + b = 79 + 40 = 119

f(x)+f(x-1/x)=1/x put x=-4 in the above equation we get f(-4)+f(5/4)=-1/4 (eqn1) put x=5/4 we get f(5/4)+f(1/5)=4/5 (eqn2) put x=1/5 we get f(1/5)+f(-4)=5 (eqn 3) now,subtract eqn2 from eqn3 and then add with eqn1. then we get 2 f(-4)=5-4/5-1/4 f(-4)=79/40 a=79,b=40 thus a+b=119

f(-4)+f(5/4)=-1/4 f(5/4)+f(1/5)=4/5 f(1/5)+f(-4)=5 f(-4)=(5-1/4-4/5)/2=79/40

x = -4

f(-4) + f(5/4) = -1/4 .....(1)

x = 5/4

f(5/4) + f(1/5) = 4/5 .....(2)

x = 1/5

f(1/5) + f(-4) = 5 ..........(3)

From 1,2 and 3

f(-4)= a/b = 79/40

So, a+b = 119

If we put $\displaystyle x=-4$ in the text, we obtain $\displaystyle f(-4)=\frac{1}{-4}-f\left(\frac{-4-1}{-4}\right)=-\frac{1}{4}-f\left(\frac{5}{4}\right)$ Puttin again $\displaystyle x=\frac{5}{4}$ in the text, we have $\displaystyle f\left(\frac{5}{4}\right)=\frac{1}{\frac{5}{4}}-f\left(\frac{\frac{5}{4}-1}{\frac{5}{4}}\right)=\frac{4}{5}-f\left(\frac{1}{5}\right)$ Again, setting $\displaystyle x= \frac{1}{5}$ in the text we obtain $\displaystyle f\left(\frac{1}{5}\right)=\frac{1}{\frac{1}{5}}-f\left(\frac{\frac{1}{5}-1}{\frac{1}{5}}\right)=5-f(-4)$ So $\displaystyle f(-4)=-\frac{1}{4}-f\left(\frac{5}{4}\right)=\\=-\frac{1}{4}-\left( \frac{4}{5}-f\left(\frac{1}{5}\right) \right)=\\=-\frac{1}{4}-\left( \frac{4}{5}-\left( 5-f(-4) \right)\right)$ that is $2f(-4)=5-\frac{1}{4}-\frac{4}{5}=\frac{100-5-16}{20}=\frac{79}{20}$ Hence we have $f(-4)=\frac{79}{40}$ and $a+b=79+40=119$ .

Given: f(x) + f(1-\frac {1}{x}) = \frac {1}{x} Putting x = -4 : f(-4) + f(\frac {5}{4}) = -(\frac {1}{4})..................(i) Putting x= \frac {5}{4} : f(\frac {5}{4}) + f(\frac {1}{5}) = \frac {4}{5}....................(ii) Putting x = \frac {1}{5} : f(\frac {1}{5}) + f(-4) = 5....................(iii) equ (ii) - equ (i) : f(\frac {1}{5}) - f(-4) = \frac {21}{20}..................(iv) equ (iii) - equ (iv) : 2f(-4) = 5 - \frac {21}{20} = \frac {79}{20} => f(-4) = \frac {79}{40} = \frac {a}{b} => a + b = 119

Letting $x=-4,\frac{4}{5},\frac{1}{5}$ gives the following equations, respectively:

$f(-4)+f(\frac{5}{4})=\frac{-1}{4}$ .

$f(\frac{5}{4})+f(\frac{1}{5})=\frac{4}{5}$ .

$f(\frac{1}{5})+f(-4)=5$ .

Adding the three equation together, we get

$2(f(-4)+f(\frac{5}{4})+f(\frac{1}{5}))=\frac{111}{20}$ .

We divide both sides by 2 to get

$f(-4)+f(\frac{5}{4})+f(\frac{1}{5})=\frac{111}{40}$ .

Subtracting this equation by the second equation from above thus yields

$f(-4)=\frac{111}{40}-\frac{4}{5}=\frac{79}{40}$ .

Our desired answer is $79+40=119$ .

Substitute $x=-4, \frac{5}{4}$ and $\frac{1}{5}$ .

$f(-4)+f(\frac{-4-1}{-4})=-\frac{1}{4}$

When simplified: $f(-4)+f(\frac{5}{4})=-\frac{1}{4}$

$f(\frac{5}{4})+f(\frac{\frac{5}{4}-1}{\frac{5}{4}})=\frac{4}{5}$

When simplified: $f(\frac{5}{4})+f(\frac{1}{5})=\frac{4}{5}$

$f(\frac{1}{5})+f(\frac{\frac{1}{5}-1}{\frac{1}{5}})=5$

When simplified: $f(\frac{1}{5})+f(-4)=5$

Adding the 3 simplified equations, we get:

$2f(-4)+2f(\frac{5}{4})+2f(\frac{1}{5})=-\frac{1}{4}+\frac{4}{5}+5=\frac{111}{20}$

Dividing by 2, we get:

$f(-4)+f(\frac{5}{4})+f(\frac{1}{5})=\frac{111}{40}$

Subtracting $f(\frac{5}{4})+f(\frac{1}{5})=\frac{4}{5}$ , we get:

$f(-4)=\frac{111}{40}-\frac{4}{5}=\frac{79}{40}$

Hence $a+b=79+40=119$ .

f(-4)=f(-4)+f( $\frac{5}{4}$ )= $\frac{-1}{4}$ --------------------------------[ $1$ ]

f( $\frac{5}{4}$ )=f( $\frac{5}{4}$ )+f( $\frac{1}{5}$ )= $\frac{4}{5}$

f( $\frac{1}{5}$ )=f( $\frac{1}{5}$ )+f( $-4$ )= $5$

f( $\frac{1}{5}$ )= $5$ -f( $-4$ )

f( $\frac{5}{4}$ )+ $5$ -f( $-4$ )= $\frac{4}{5}$

f( $\frac{5}{4}$ )= $\frac{-21}{5}$ +f( $-4$ )

Putting value of f( $\frac{5}{4}$ ) in equation [ $1$ ]

we get

f( $-4$ )+f( $-4$ )- $\frac{-21}{5}$ = $\frac{-1}{4}$

On solving we get

f( $-4$ )= $\frac{79}{40}$

Thus $a$ = $79$ $and$ $b$ = $40$

thus $a$ + $b$ = $\boxed{119}$

We see that setting $x=\,-4,\,f(-4)+f\Big(\displaystyle\frac{5}{4}\Big)=\displaystyle\frac{-1}{4}\;\;\;\;---(i)$

setting $x=\displaystyle\frac{5}{4},\,f\Big(\displaystyle\frac{5}{4}\Big)+f\Big(\displaystyle\frac{1}{5}\Big)=5\;\;\;---(ii)$

setting $x=\displaystyle\frac{1}{5},\,f\Big(\displaystyle\frac{1}{5}\Big)+f(-4)=5\;\;\;---(iii)$ .

Now $,(i)-(ii)+(iii)$ gives $f(-4)=\displaystyle\frac{79}{40}=\displaystyle\frac{a}{b}.$

Thus our required answer is $a+b=79+40=\boxed{119}.$

first, we look at the function f( \frac {x-1}{x} ). we can see that it's equal to f( 1- \frac {1}{x} )

we put x= -4 then f(-4) + f(1 - /frac {1}{-4}) = /frac {1}{-4}

then f(-4) + f(/frac {5}{4} ) = /frac {-1}{4}

by applying the functiun to x= /frac {5}{4}

we have: f(/frac {5}{4}) + f( /frac {1}{5}) = /frac {4}{5}

and by applying it to x= /frac {1}{5}

we have f( /frac {1}{5}) + f(-4)= 5 thus, we have 3 different functions and 3 equations, so we put:

x= f(-4)

y= /frac {5}{4}

z= /frac {1}{5}

we get

x+y= /frac {-1}{4}

y+z= /frac {4}{5}

z+x= 5

by using the 3rd equality we have : z= 5-x

by replacing it in the second equality we get:

y+5-x= /frac {4}{5}

thus y-x= /frac {-21}{5}

and we have: x+y= /frac {-1}{4}

by adding them side by side we get

2y= /frac {1}{4} +/frac {-21}{5}

2y= /frac {-89}{20}

so y= /frac{-89}{40}

and we have: x+y= /frac {-1}{4}

then x= /frac {-1}{4} - /frac{-89}{40}

x= /frac{79}{40} then f(-4) = /frac{79}{40}

since 79 and 40 are coprime, and f(-4)= /frac{a}{b} where a and b are coprime, then a=79 and b=40, thus a+b=79+40=119

Given ; $f(x) + f\left( \frac {x-1} {x} \right) = \frac{1}{x} ...(1)$ .Make the transformation $x \large \rightarrow$ $1 - \dfrac{1}{x}$ Now we will get

$f\left(1-\dfrac{1}{x}\right) + f\left( \frac {1} {1-x} \right) = \frac{x}{x-1} ... (2)$

Again make the transformation $x \large \rightarrow$ $1 - \dfrac{1}{x}$ We will get

$f\left(\dfrac{1}{1-x}\right) + f\left( x \right) = 1-x ... (3)$

$(1) - (2) + (3)$ gives,

$\large {\boxed{ f(x) = \frac{x^3-x^2+1}{2x(1-x)} }}$