How do you graph $y = {tan}^{- 1} (2 x - 4)$ $y=tan^-1(2x-4)$ over the interval $- 2 \leq x \leq 6$ $-2<=x<=6$ ?

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Answer 1 · 2017-02-03T05:40:32

Graphs are gifts, from the Socratic utility.

$y = {tan}^{- 1} (2 x - 4)$ $y = tan^(-1)(2x-4)$

is inversely ( for the same graph but with restricted

$y \in (- \frac{π}{2}, \frac{π}{2})$ $y in (-pi/2, pi/2)$ )

$x = 2 + (\frac{1}{2}) tan y$ $x=2 + (1/2) tan y$

Direct graph, using $y = arctan (2 x - 4)$ $y = arctan(2x-4)$ :
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

$x = 2 + (\frac{1}{2}) tan y$ $x = 2 +(1/2) tan y$ , for $y \in (- \frac{π}{2}, \frac{π}{2})$ $y in (-pi/2, pi/2)$ . Of course, this includes

ranges $(- \infty, \infty)$ $(-oo, oo)$ 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 $y = k π + {tan}^{- 1} (2 x - 4)$ $y = kpi + tan^(-1)(2x-4)$ , k =

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

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