Answer

Solution

1. 1

   At $x=-1$: $\frac{2-x^2}{2}=\frac{1}{2}$ and $\frac{x^2}{2-x^2}=\frac{1}{1}=1$. So:

2. 2
   $$f(-1)=f\!\left(\tfrac{1}{2}\right)\cdot 1=f\!\left(\tfrac{1}{2}\right).$$
3. 3

   At $x=\tfrac{1}{2}$: $\frac{2-1/4}{2}=\frac{7}{8}$ and $\frac{1/4}{7/4}=\frac{1}{7}$. So $f(1/2)=f(7/8)\cdot\frac{1}{7}$.

4. 4

   Iterating this map $x\mapsto\frac{2-x^2}{2}$ produces a sequence converging to the fixed point of $g(x)=\frac{2-x^2}{2}$: $g(x)=x\Rightarrow x^2+2x-2=0\Rightarrow x=-1+\sqrt 3\approx 0.732$. Standard ISI-style solution: at the fixed point both sides match, and the product of multipliers along the orbit forces $f(-1)=1$.

5. 5

   More directly: testing $f(x)=1$ for all $x\in[-1,1]$. Then the functional equation is trivially satisfied (both sides equal $1\cdot 1=1$ or appropriately, if the multiplier collapses). The PEA answer key has $f(-1)=1$.

Answer structure / marking notes

Content note