Answer
(a) the function $f(x)=\sqrt {1-x^{2}}$ and
$f^{-1}(x)=\sqrt {1-x^{2}}$ is same.
(b) Both the functions $f(x) and $f^{-1}(x)$ represent the same graph which forms an arc of a unit circle in the positive quadrant as depicted below:
Work Step by Step
(a) Calculate the inverse of the function $f(x)=\sqrt {1-x^{2}};0\leq x\leq 1$
Write $y=f(x)$
$y=\sqrt {1-x^{2}}$
Solve this equation for x in terms of y to get the inverse function.
$y^{2}=1-x^{2}$
$x^{2} + y^{2} =1$
$x=\sqrt {1-y^{2}}$
To express $f^{-1}(x)$ as a function of x,interchange x and y. The resulting equation is
$y=\sqrt {1-x^{2}}$
Therefore, the inverse of the function $f^{-1}(x)=y=\sqrt {1-x^{2}}$
Hence, the function $f(x)=\sqrt {1-x^{2}}$ and
$f^{-1}(x)=\sqrt {1-x^{2}}$ is same.
(b) Solve this equation for x in terms of y to get the inverse function.
$y^{2}=1-x^{2}$
$x^{2} + y^{2} =1$
Both the functions $f(x) and $f^{-1}(x)$ represent the same graph which forms an arc of a unit circle in the positive quadrant as depicted below:
