# Solve the equation fof(x)=0 Where f(x)=1/(x^2+x)

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### 3 Answers

We have to findx given that fof(x)=0 and f(x)=1/(x^2+x).

fof(x) = f(f(x))

=> f(1 / (x^2 + x))

=> 1/((1/ x^2 + x)^2 + (1/ x^2 + x))

=> (x^2 + x)^2 / (x^2 + x + 1)

Now this is equal to zero

=> (x^2 + x)^2 / (x^2 + x + 1) = 0

=> (x^2 + x)^2 = 0

=> x^2( x + 1)^2 = 0

So we get x = 0 and x = -1

**Therefore x= 0 and x = -1.**

f(x) = 1/(x^2+x). To solve the equation fof(x) = =.

fof(x) = 1/{(f(x))^2+f(x)} = 0.

1/{(f(x))^2+f(x)} = 1/{(1/(x^2+x))^2 + 1/(x^2+x)} = (x^2+x)^2/{1+(x^2+x)} = x^2(x+1)^2/{x^2+x+1).

fof(x) = 0

=> x^2(x+1)^2/{x^2+x+1) = 0. We multiply by x^2+x+1 and get:

=> x^2(x+1)^2 = 0.

So x^2 = 0, or (x+1)^2 = 0.

x= 0, or x=-1.

Therefore the solution of fof(x) = 0 is x= 0, or x= -1.

(fof)(x) = f(f(x))

We'll substitute x by f(x) and we'll get:

f(f(x)) = 1/[f^2(x)+f(x)]

f(f(x)) = 1/[1/(x^2+x)^2 + 1/(x^2+x)]

f(f(x)) = (x^2+x)^2/(1 + x + x^2)

We'll put f(f(x)) = 0:

(x^2+x)^2/(1 + x + x^2) = 0

Since the denominator is always positive, we'll put the numerator as zero:

(x^2+x)^2 = (x^2+x)(x^2+x)

(x^2+x)^2 = 0

(x^2+x)(x^2+x) = 0

We'll factorize by x both brackets:

x*x(x+1)(x+1) = 0

**x1 = 0 and x2 = 0**

x + 1 = 0

**x3 = -1 and x4 = -1**

**The solutions of the equation (fof)(x) = 0 are {-1 ; 0}.**