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.