Let r be a real number and $f(x) = \begin{cases} 2x-r & \text{if } x \ge r \\ 1 & \text{if } x < r \end{cases}$. Then, the equation $f(x) = f(f(x))$ holds for all real values of x where.

Let's solve this by checking when $f(x) = f(f(x))$ works for different values of $x$.

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Our piecewise function is:

• $f(x) = 2x - r$ when $x \geq r$
• $f(x) = 1$ when $x < r$

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When $x = r$:

$f(r) = 2r - r = r$ (since $r \geq r$, we use the first piece)

$f(f(r)) = f(r) = r$

So $f(r) = f(f(r))$

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When $x < r$:

$f(x) = 1$ (using the second piece)

Now we need $f(f(x)) = f(1)$

We need to determine whether $1 \geq r$ or $1 < r$:

If $1 < r$: $f(1) = 1$

So $f(f(x)) = f(1) = 1 = f(x)$

If $1 \geq r$: $f(1) = 2(1) - r = 2 - r$

So $f(f(x)) = 2 - r$, but $f(x) = 1$

For this to work: $1 = 2 - r$, which gives $r = 1$

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When $x > r$:

$f(x) = 2x - r$

Since $x > r$, we have $2x - r > 2r - r = r$

So $f(f(x)) = f(2x - r) = 2(2x - r) - r = 4x - 3r$

For $f(x) = f(f(x))$: $2x - r = 4x - 3r$

Solving: $2r = 2x$, so $x = r$

But this contradicts our assumption that $x > r$!

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The equation $f(x) = f(f(x))$ holds when:

• $x = r$ (always works)
• $x < r$ (works when $r \geq 1$)
• $x > r$ (never works)

Therefore, the equation $f(x) = f(f(x))$ holds for all real values of $x$ where $x \leq r$ and $r \geq 1$.