## College Algebra (6th Edition)

$a)$ $f^{-1}(x)=\dfrac{1}{x}$ $b)$ $f(f^{-1}(x))=f^{-1}(f(x))=x$
$f(x)=\dfrac{1}{x}$ $a)$ Substitute $f(x)$ by $y$: $y=\dfrac{1}{x}$ Interchange $x$ and $y$: $x=\dfrac{1}{y}$ Solve for $y$: $xy=1$ $y=\dfrac{1}{x}$ Substitute $y$ by $f^{-1}(x)$: $f^{-1}(x)=\dfrac{1}{x}$ $b)$ Verify the equation for the inverse function found by finding $f(f^{-1}(x))$. Substitute $x$ by $f^{1}(x)$ in $f(x)$ and simplify: $f(f^{-1}(x))=\dfrac{1}{\Big(\dfrac{1}{x}\Big)}=x$ Complete the verification process by finding $f^{-1}(f(x))$. Substitute $x$ by $f(x)$ in $f^{-1}(x)$ and simplify: $f^{-1}(f(x))=\dfrac{1}{\Big(\dfrac{1}{x}\Big)}=x$ Since $f(f^{-1}(x))=f^{-1}(f(x))=x$, the equation for the inverse function found is correct.