Solutions to a Recursive Function

We first note that if $x = a \ne 0$ , we cannot have $f(a) = 0$ . This is simple: suppose that $f(a) = 0$ for $a \ne 0$ from which it follows directly that $a =f(f(a)) = f(0)$ . However, substituting $x =0$ gives $f(0) = f(f(0))$ , or $a = f(a) = 0$ , which is a contradiction to our supposition that $a \ne 0$ .

Now, we show that $x=0$ must be a solution to $f(x) = 0$ . Suppose that $f(0) = a \ne 0$ . Then, $0 + a = f(a)$ , but then $a+f(a) = f(f(a)) = f(a)$ , so $a=0$ , and once again we have a contradiction. Hence, $x=0$ is the one and only solution to $f(x) = 0$ .

First, we will prove that $f(x)$ is an injective function. What we have to prove is that for any $x_1,x_2\in\mathbb{R}$ it holds that ${f(x_1)=f(x_2)\iff{x_1=x_2}}$ . We proceed as follows:

$f(x_1)=f(x_2)\Rightarrow f(f(x_1))=f(f(x_2))\Rightarrow f(x_1)+x_1=f(x_2)+x_2\Rightarrow x_1=x_2$ Since $f$ is not a multivalued function it is also true that $x_1=x_2\Rightarrow{f(x_1)=f(x_2)}$ . Now that $f$ is injective there can be no more than one solution to the equation $f(x)=0$ in the function's domain.

Setting $x=0$ in the recursive relationship we obtain: $f(f(0))=f(0)\Leftrightarrow$ $f(0)=0$ . The former holds because $f$ has been proven to be injective. So $x=0$ is a solution to $f(x)=0$ and obviously the only one.

By substituting $x=0$ , we get $f(0)=f(f(0))$ . By substituting $x= f(0)$ , we get $f(0)+f(f(0))=f(f(f(0)))$ . But since $f(0)= f(f(0))$ , the equation above is actually equivalent to $f(0)+f(f(0))=f(f(0))$ . This implies that $f(0)=0$ . Hence, whe know there exists at least one $x$ such that $f(x)=0$ . By substituting $f(x)=0$ into our original equation, we get $f(0)=x$ . Therefore, since $f(0)$ is a constant, there is only one possible $x$ (which is the value 0), hence the answer is 1. note: the interval $[-10,10]$ is not necessary.

[Latex Edits - Calvin]

Substitute $x=0$ for the functional equation and get $0+f(0)=f(f(0))$ so $f(0)=f(f(0))$ .

Substitute $f(0)$ for the functional equation and get $f(0)+f(f(0))=f(f(f(0)))$ so $2f(0)=f(0)$ . Hence, $f(0)=0$ is one solution.

Suppose there is a solution $a$ . Then $f(a)=0$ . Substituting $a$ into the functional equation yields $a+f(a)=f(f(a))$ so $a+0=f(0)=0$ . Hence, $a=0$ so $f(0)=0$ is the only solution.

Let $x=c$ be a solution to $f(x)=0$ .

Using the property of the function as stated in the problem, we get $c+f(c)=f(f(c))$ . But since the real number $c$ is a solution to $f(x)=0$ , the previous equation will be reduced to $c=f(0)$ . Since $f$ is defined on all reals, $f(0)$ exists and is unique.

If $x_0$ is a solution, then $f(x_0)=0$ . Hence $x_0+f(x_0)=f(f(x_0))$ which gives $x_0=f(0)$ . Since $f(0)$ is unique it follows that $x_0$ is unique, qed.

f(x)=0 <=> f(0)=x, such an x exists by definition.

If for some x,y f(x)=f(y), x+f(x)=f(f(x))=f(f(y))=f(y)+y <=> x=y Thus, there can only be at most x such that f(x)=0.

Thus there is 1 root to the function.

in the equation x + f(x) = f(f(x)), put x=0 to get f(0) = f(f(0)). Now put x= f(0) in the equation to get 2 f(0) = f(0) and so f(0)=0. Let f(a)=0. Then, put x=a to get a = f(0) = 0. Thus, the unique solution of f(x)=0 is x=0. hence answer is 1.

Let $f(z)=0$ for some real $z$ . We prove that $z$ must be $0$ .

For this $z$ , $z=z+f(z)=f(f(z))=f(0)$ . Plug $x=0$ into it yields $z=f(0)=f(f(0))=f(z)=0$ .

Conversely, we show that $f(0)=0$ . Let $f(0)=y$ , then $y=0+f(0)=(f(0))=f(y)$ . Then let $x=y$ we know that $2y=y+y=y+f(y)=f(f(y))=f(y)=y$ , so $y=0$ . Therefore in any function satisfy the relation above, $f(x)=0\Leftrightarrow x=0$ . The answer is 1.

Let f(a)=0. Then a + f(a) = f(f(a)) implies that a = f(0). Since f(x) is a function, there is a value of a such that a=f(0). Therefore, f(x) has exactly one root. It suffices to show that this root is in the interval [-10,10]. Note that plugging in x=0 into the functional equation implies that 0 + f(0) = f(f(0) = f(a) = 0. Therefore, f(x) has a root at x=0, which is contained within the given interval.

x+f(x)=f(f(x))and f(x)=0,implies x=f(0).As the function is from reals to reals there exists only one solution for x=f(0)

First, we will show that the function is injective. If $f(x) = f(y)$ , then $f(f(x)) = f( f(y))$ , so $x = f(f(x)) - f(x) = f( f(y) ) - f(y) = y$ .

Since $f( f( 0 ) ) = f(0) + 0 = f(0)$ , by injectivity we must have $f(0) = 0$ . If there is a number $x$ such that $f(x) = 0 = f(0)$ , then by injectivity $x=0$ . Hence, the only solution in the interval is $x=0$ . Hence, there is 1 solution.

f(x) must be a linear polynomial. iIf f(x) is polyinomial with degree n then the degree of f(f(x)) will be n^2 but x+f(x)=f(f(x)) which means that f(x) must have the same degree of f(f(x)) i.e n=n^2. so n=2 or 0 cannot be = 0 so n=1 therefore f(x) is a linear polynomial , it is clear that f(x)=ax+b x+ax+b=a(ax+b)+b ab+b=b ab=0 a cannot be equal to 0 therefore b=0 f(x)=ax from the given question, x+f(x)=f(f(x)) x+ax=(a^2)x by solving for a you get a =(1 +/- √5)/2 f(x)=x(1 +/- √5)/2 f(x)=0 only when x=0 there is only one solution

it is given that x+f(x) =f((x)) and f(x) =0, where f is a function satisfying all real values.as it is a function each domain has only one co domain.

then by substituting the value of f(x) in the given equation we get, x+0 =f(0) or f(0) is x so as it is a function it has only one value of x. hence the answer will be 1.

it is given that f(x) =0 where f is a function satisfying all real values but x+f(x) =f((x))

so by substituting the value of f(x) in the given relation we get, x+0 =f(0) or f(0) = x as f is a real to real function so there will be only one possible value of x. hence answer is 1

IF WE SUBSTITUTE 0 IN THE GIVEN EQUATION,IT WILL ONLY SATISFY THE EQUATION IN THE GIVEN INTERVAL

Without loss of generality, Let us assume f(0) = a ---- equation 1 By the given condition,
f(f(0)) = f(0) + 0 substituting from equation 1 f(a) = a --- 2 So, f(f(a)) = f(a) + a f(a) = 2a ---- 3 Since the function cannot have 2 values simultaneously, (from 2 and 3) 2a=a => a=0 therefore f(0) = 0 Let any other value 'x' have f(x) = 0 So, f(f(x)) = f(x) + x => f(0) = 0 + x {but f(0)=0} so, x=0 which is our original solution. Therefore, f(0) = 0 is the only possibility Ans: One Solution.

If $f\left(x\right)=0$ , $x+f\left(x\right)=x+0=x=f\left(x\right)$ . Again, we know that $f\left(x\right)=0$ so this can only occur at $x=f\left(x\right)=0$ . $0$ .