How do you graph #y=tan^-1(2x-4)# over the interval #-2<=x<=6#?

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Answer 1 · 2017-02-03T05:40:32

Graphs are gifts, from the Socratic utility.

#y = tan^(-1)(2x-4)#

is inversely ( for the same graph but with restricted

#y in (-pi/2, pi/2)#)

#x=2 + (1/2) tan y#

Direct graph, using #y = arctan(2x-4)#:
graph{(y-arctan(2x-4))(y+1.573+.01x)(y-1.573+.01x)=0 [-2 6 -1.8 1.8]}

Same graph, using the inverse

#x = 2 +(1/2) tan y#, for #y in (-pi/2, pi/2)#. Of course, this includes

ranges #(-oo, oo)# for both x and y. Slide the cursor over the graph

#uarr# and #darr# to see the extended graph for (one x, many y)

plots, from the reverse inverse #y = kpi + tan^(-1)(2x-4)#, k =

integer.
graph{(x-2-0.5 tan y )(y+1.573+.01x)(y-1.573+.01x)=0 [-2 6 -1.8 1.8]}