Select the [b]Default to Natural Domain of f [/b]checkbox. Then select [b][color=#cc0000]Show Inverse Relation[/color]. [/b]Is this [b][color=#cc0000]inverse relation [/color][/b](whose equation is [math]x=cos\left(y\right)[/math] ) also a function? Explain why or why not.
No, it is not. In order for a relation to be a function, each input can have only one output. Yet in order for the [color=#cc0000][b]inverse relation[/b][/color] to be a function, [b][color=#1e84cc]each output [/color][/b](of the [color=#bf9000][b]original function[/b][/color]) can only have 1 input that maps to it. Yet this clearly isn't the case for the function [math]y=cos\left(x\right)[/math] because the [color=#1e84cc][b]output[/b][/color] 1/2, for example, has more than one input (angle) (ex's: [math]\frac{\pi}{3}[/math] and [math]\frac{5\pi}{3}[/math] ) that map to it. (In fact, there exist infinitely many angles whose cosine ratio is 1/2.) Therefore, the [color=#cc0000][b]inverse relation[/b][/color] is not a function.