How do you show that $f (x) = 9 - x^{2}$ $f(x)=9-x^2$ and $g (x) = \sqrt{9 - x}$ $g(x)=sqrt(9-x)$ are inverse functions algebraically and graphically?

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Answer 1 · 2017-01-31T17:54:30

See proof below

We calculate the composition of the functions

$f (x) = 9 - x^{2}$ $f(x)=9-x^2$

$g (x) = \sqrt{9 - x}$ $g(x)=sqrt(9-x)$

$f (g (x)) = f (\sqrt{9 - x}) = 9 - {(\sqrt{9 - x})}^{2}$ $f(g(x))=f(sqrt(9-x))=9-(sqrt(9-x))^2$

$= 9 - (9 - x) = x$ $=9-(9-x)=x$

$g (f (x)) = g (9 - x^{2}) = \sqrt{9 - (9 - x^{2})}$ $g(f(x))=g(9-x^2)=sqrt(9-(9-x^2))$

${\sqrt{x}}^{2} = x$ $sqrtx^2=x$

Therefore, $f (x)$ $f(x)$ and $g (x)$ $g(x)$ are inverses

$f (x) = g^{- 1} (x)$ $f(x)=g^-1(x)$

and $g (x) = f^{- 1} (x)$ $g(x)=f^-1(x)$

Graphically, $f (x)$ $f(x)$ and $g (x)$ $g(x)$ are symmetrics wrt $y = x$ $y=x$

graph{(y-9+x^2)(y-sqrt(9-x))(y-x)=0 [-1.78, 43.84, -9.37, 13.44]}

How do you show that f(x)=9−x2f(x)=9-x^2 and g(x)=√9−xg(x)=sqrt(9-x) are inverse functions algebraically and graphically?