#### Answer

a) $f^{-1}(x)=\sqrt{1-x^2}=f(x)$
The inverse is the same as the original function.
b) The graph of $f$ is a quarter of a circle with centre at the origin and radius $1$(u) located in the first quadrant.
Since $f$ is symmetrical about $y=x$, its inverse is identical to itself.

#### Work Step by Step

a) $f(x)=y=\sqrt{1-x^2}, 0\leq x\leq1$
$y^2=1-x^2$
$x^2=1-y^2$
$x=\pm\sqrt{1-y^2}\geq0$
$\therefore x=\sqrt{1-y^2}$
$y\rightleftharpoons x$
$f^{-1}(x)=y=\sqrt{1-x^2}=f(x)$
The inverse is the same as the original function.
b) $f(x)=y=\sqrt{1-x^2}$
Domain: $0\leq x\leq1$
Range: $0\leq y\leq1$
$\therefore$Graph exists only in 1st quadrant
Squaring both sides: $y^2=1-x^2$
$(x-0)^2+(y-0)^2=1^2$
Circle with centre $(0,0)$, r=$1$(u)
The graph of $f$ is a quarter of a circle with centre at the origin and radius $1$(u) located in the first quadrant.
Since $f$ is symmetrical about $y=x$, its inverse is identical to itself.