## Calculus: Early Transcendentals (2nd Edition)

$a.$ $y=f^{-1}(x)=\dfrac{6-x}{4}$ $b.$ The relationships $f^{-1}(f(x))=f(f^{-1}(x))=x$ are verified.
$f(x)=6-4x$ $a.$ Substitute $f(x)$ by $y$: $y=6-4x$ Solve the equation for $x$. Begin by taking $4x$ to the left side and $y$ to the right side: $4x=6-y$ Take the $4$ to divide the right side: $x=\dfrac{6-y}{4}$ Interchange $x$ and $y$ and write the inverse in $y=f^{-1}(x)$ form: $y=f^{-1}(x)=\dfrac{6-x}{4}$ $b.$ Check the inverse found by evaluating $f(f^{-1}(x))$ and $f^{-1}(f(x))$: $f(f^{-1}(x))=6-4\Big(\dfrac{6-x}{4}\Big)=6-6+x=x$ $f^{-1}(f(x))=\dfrac{6-(6-4x)}{4}=\dfrac{6-6+4x}{4}=\dfrac{4x}{4}=x$ The relationships $f^{-1}(f(x))=f(f^{-1}(x))=x$ are verified.