3: Given the function $f(x)=\sqrt{1-x}(x \leqslant 1)$. Then the coordinates of the intersection point of the function $f(x)$ and its inverse function $f^{-1}(x)$ are $\qquad$
3. $(1,0),(0,1),\left(\frac{-1+\sqrt{5}}{2}, \frac{-1+\sqrt{5}}{2}\right)$.
The inverse function of $f(x)=\sqrt{1-x}(x \leqslant 1)$ is $f^{-1}(x)=1-x^{2}(x \geqslant 0)$.
Solve the system of equations
$$
\left\{\begin{array}{l}
y=\sqrt{1-x}, \\
y=1-x^{2} .
\end{array}\right.
$$

From equation (1), we get
$$
y^{4}=1-2 x+x^{2}=2(1-x)-\left(1-x^{2}\right) \text {. }
$$

Substituting equations (1) and (2) into the above equation, we get $y^{4}=2 y^{2}-y$, which simplifies to
$$
\left(y^{4}-y^{2}\right)-\left(y^{2}-y\right)=0 \text {, }
$$

Thus, $y(y-1)\left(y^{2}+y-1\right)=0$.
Solving this, we get $y_{1}=0, x_{1}=1 ; y_{2}=1, x_{2}=0$; $y_{3}=\frac{-1+\sqrt{5}}{2}$ (discarding the negative value), $x_{3}=\frac{-1+\sqrt{5}}{2}$.