Let f(x)=x^2 -1 . How many distinct real roots are there to f(f(f(x)))
To determine the number of distinct real roots of the function f(f(f(x))), we need to find the composition of f with itself three times.
First, let's calculate f(x):
f(x) = x^2 - 1
Next, let's calculate f(f(x)):
f(f(x)) = f(x^2 - 1) = (x^2 - 1)^2 - 1 = x^4 - 2x^2 + 1 - 1 = x^4 - 2x^2
Finally, let's calculate f(f(f(x))):
f(f(f(x))) = f(f(x^4 - 2x^2)) = ((x^4 - 2x^2)^2 - 1)^2 - 1
To find the distinct real roots, we need to solve the equation f(f(f(x))) = 0.
((x^4 - 2x^2)^2 - 1)^2 - 1 = 0
Simplifying further, let's set u = x^2:
((u^2 - 2u)^2 - 1)^2 - 1 = 0
Now we have a quartic equation in terms of u, which we can solve to find the values of u.
(u^2 - 2u)^2 - 1 = ±1
(u^2 - 2u)^2 = 2
Taking the square root:
u^2 - 2u = ±√2
Rearranging the equation:
(u - 1)^2 = ±√2 + 1
(u - 1)^2 = √2 + 1 or (u - 1)^2 = -√2 - 1
Solving each equation separately:
For (u - 1)^2 = √2 + 1:
u - 1 = ±√(√2 + 1)
Solving for u:
u = 1 ± √(√2 + 1)
Since u = x^2, we must consider the solutions for x:
x = ±√(1 ± √(√2 + 1))
Similarly, for (u - 1)^2 = -√2 - 1:
u - 1 = ±i√(1 + √2)
This equation has no real solutions, so we can ignore it in terms of finding distinct real roots.
Therefore, the function f(f(f(x))) has two possible distinct real roots given by:
x = ±√(1 ± √(√2 + 1))
So, the function f(f(f(x))) has two distinct real roots.