Graph the function f(x)=3/2x-arccos(1/x).

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Geometer's Sketchpad shows the graph is discontinous from -1 to +1.  So the straight line should be eliminated.

As justaguide stated The graph of f(x) = arccos(x) has a domain of [-1, 1] as the sine of any number lies between -1 and 1. Therefore F(x)=arccos 1/x has a domain of x<-1 or x>1, it is undefined between -1 and 1 as that would be the arccos of a number greater than 1.

 

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The graph of f(x) = arccos(x) is as shown below. arccos(x) has a domain of [-1, 1] as the sine of any number lies between -1 and 1.

(3/2x) can take on a value for any x except x = 0 which makes 3/2x  undefined.

Subtracting arccos(1/x) from 3/2x gives the following graph

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