## Calculus (3rd Edition)

$$y=-\frac{1}{2} x+6$$
Since $f(x)$ has the tangent line $y=-2x+12$ at $x=4$, then $f(x)$ passes through $$(4,-2(4)+12)=(4,4)$$ Since $g(x)$ is the inverse of $f(x)$, and $f, g$ are symmetric at $x=y$, then \begin{align*} g^{\prime}(4)&=\frac{1}{f^{\prime}(g(4))}\\ &=\frac{1}{f^{\prime}(4)}\\ &=\frac{1}{-2} \end{align*} Hence, the tangent line of $g(x)$ at $(4,4)$ is given by \begin{align*} \frac{y-y_1}{x-x_1}&=m\\ \frac{y-4}{x-4}&=\frac{-1}{2} y&= -\frac{1}{2} x+6 \end{align*}