
Kenneth S. answered 06/15/17
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Expert Help in Algebra/Trig/(Pre)calculus to Guarantee Success in 2018
Interchange variables, i.e. x = 4y2 + 1.
Solve for the y! Relabel it as the inverse of f!
In this instance, the quadratic function f does not have an inverse that is also a function, for two reasons:
(1) when you attempt to solve for the new y, you have to take square root and put ± in front of that expression, so for some x, you get TWO y values,
(2) original graph fails the HORIZONTAL LINE TEST, which is necessary for f to have an inverse that is also a function.