How do you graph -2 tan^(-1) (x/4)−2tan−1(x4)?

1 Answer
A. S. Adikesavan
Jan 4, 2017

Graphs are inserted and explained.

Explanation:

I have now adopted the convention tan^(-1)theta in (-pi/2, pi/2)tan−1θ∈(−π2,π2) only.

y=-2tan^(-1)(x/4) in -2(-pi/2, pi/2)=-(-pi, pi)y=−2tan−1(x4)∈−2(−π2,π2)=−(−π,π),

giving 1 - 1 relation.

Inversely,

x = -4 tan(y/2),. y in (-pi. pi) and x in (-oo, oo)x=−4tan(y2),.y∈(−π.π)andx∈(−∞,∞).

In the inserted graph, the required graph is restricted to

y in (-pi. pi)y∈(−π.π).

This covers one y-period (pi)/(1/2) = 2piπ12=2π. The terminal asymptotes

are y=+-piy=±π.

Graph of y = -2 tan^(-1) (x/4)y=−2tan−1(x4):

graph{x+4tan(y/2)=0[-500 500 -3.4 3.1416]}

Breaking the conventional 1 - 1 rule, 'one x - many y' graph is also presented. The terminal asymptotes are y = +-k pi, k = 1, 2, 3. 4, .... See below.

graph{x+4tan(y/2)=0[-500 500 -20 20]}

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