To find $f(2)$, I need to know the slope and intercept of the linear function $f(x)$. I can use the given equation and the fact that $f(1) = 4$ to find both of these values.
[*  I can start by plugging in $x = 1$ into the given equation and see what happens. I get $f(1) = 4f^{-1}(1) + 6$, which implies that $4 = 4f^{-1}(1) + 6$. Solving for $f^{-1}(1)$, I get $f^{-1}(1) = -\frac{1}{2}$. This means that $f(-\frac{1}{2}) = 1$, since $f$ and $f^{-1}$ are inverses of each other. Now I have two points on the line $f(x)$: $(1,4)$ and $(-\frac{1}{2},1)$. I can use these points to find the slope of the line by using the formula $m = \frac{y_2 - y_1}{x_2 - x_1}$. Plugging in the coordinates, I get $m = \frac{1 - 4}{-\frac{1}{2} - 1} = \frac{-3}{-\frac{3}{2}} = 2$. So the slope of $f(x)$ is $2$. To find the intercept of $f(x)$, I can use the point-slope form of a line: $y - y_1 = m(x - x_1)$. Plugging in any of the two points and the slope, I get $y - 4 = 2(x - 1)$. Simplifying, I get $y = 2x + 2$. So the intercept of $f(x)$ is $2$. Now I have the slope-intercept form of $f(x)$: $f(x) = 2x + 2$. To find $f(2)$, I just need to plug in $x = 2$ into this equation. I get $f(2) = 2(2) + 2 = 6$.


Suppose that $f(x)$ is a linear function satisfying the equation $f(x) = 4f^{-1}(x) + 6$. Given that $f(1) = 4$, find $f(2)$.