It looks like you want to find the inverse of f(x)...(not the "inverse derivative" )....this one is not easy.....
WolframAlpha gives the inverse as....
To evaluate f-1(3).....we can evaluate f(x) where y= 3
So
3 = (x^3)/4 +x - 1 .... when x =2 .....see this graph......https://www.desmos.com/calculator/qshuoxun0d
So, if (2, 3) is on f(x) then (3, 2) is on the inverse.....thus f-1(3) = 2
I beleve this person wants the cross-product ....(0, 11, 2) x (0, 5, 3)
This will form a vector that is perpendicular to both of these
To get the x coordinate, we have
(11* 3) - (2 * 5) = 33 - 10 = 23
To get the y coordinate, we have
(2*0) - (0* 3) = 0
To get the z coordinate, we have
(0 * 5) - (11* 0) = 0
So....our cross-product vector is (23, 0, 0)
This makes sense......since both of the given vectors lie in the y-z plane......the vector perpendicular to both should just lie along the x axis....!!!
Here's a pic.....
Yuo can read more about this proceedure, here ......http://www.mathsisfun.com/algebra/vectors-cross-product.html