## Calculus: Early Transcendentals 8th Edition

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.
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.