This question is about finding the slope of the tangent line to an inverse function. I would love to know if my answer of 1/5 is correct.g(1) = 2g^-1(2) = 1(g^-1(x))' = 1/(g'(g-1(x))(g^-1(2))' = 1/g'(g-1(2))(g^-1(2))' = 1/g'(1)g(x) = x^3 + 2x - 1g'(x) = 3x^2 + 2g'(1) = 3+2=5(g^-1(2))' = 1/5

Yes choice B) 1/5 is correct. You are using the correct formula and steps. Nice work.

Here's another way to look at it:

When you visually plot the inverse, you reflect everything over the line y = x. What happens is that every (x,y) point swaps coordinates. So if (1,2) is on g(x), then (2,1) is on the inverse, and vice versa. Swapping the x and y values will swap the nature of the tangent slope.

For any tangent on g(x), its slope is m = (dy)/(dx)

The corresponding slope on the inverse will be n = (dx)/(dy). All that's happened here is that the numerator and denominator flipped. Again because x and y have swapped places. If you wanted, you can picture the xy axis where the vertical axis is the x axis and the horizontal axis is the y axis. That's effectively what is going on here. 

So before you found that g'(1) = 5 meaning that m = 5/1 on g(x) at the point (1,2)
So n = 1/m = 1/5 for the point (2,1) on the inverse g function.