How do you sketch the graph of $f (x) = arctan (2 x - 3)$ $f(x)=arctan(2x-3)$ ?

Question

3162 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-26T02:27:25

See graph and details.

$y = arctan (2 x - 3) \in (- \frac{π}{2}, \frac{π}{2})$ $y = arctan ( 2 x - 3 ) in ( -pi/2, pi/2 )$

$y \neq$ $y ne$ asymptotuc $\pm \frac{π}{2}$ $+-pi/2$

Inversely,

$x = \frac{1}{2} (tan y + 3) \Rightarrow$ $x = 1/2 ( tan y + 3 ) rArr$ the period in y-directions = $π$ $pi$ .

So, one period is $y \in (\frac{π}{2}, \frac{π}{2})$ $y in ( pi/2, pi/2 )$ , the range of y..

See graph that is restricted to one period:
graph{(y - arctan(2x-3))(y^2-1/4(pi)^2)=0}

For the piecewise-wholesome inverse

$y = {(tan)}^{- 1} (2 x - 3) = k π$ $y = (tan)^(-1) ( 2x - 3 ) = kpi$ + given $y, k = 0, \pm 1, \pm 2, \pm 3,$ $y, k = 0, +-1, +-2, +-3,$ .

the graph is immediate, using the inverse of the given equation

$x = \frac{1}{2} (tan y + 3)$ $x = 1/2 ( tan y + 3 )$ .

See graph.
graph{x-1/2(3 + tany)=0[-20 20 -10 10]}
It is a wrong practice to swap ( x, y ) to ( y, x ) and call

$y = \frac{1}{2} (tan x + 3)$ $y = 1/2 ( tan x + 3 )$

the inverse of the given equation, and rotate the graph for the

given equation, for $y \in (- \frac{π}{2}, \frac{π}{2})$ $y in ( - pi/2, pi/2 )$ .

In te chosen piece, the graphs of both the given equation and its

wholesome inverse ought to be the same. ..

How do you sketch the graph of f(x)=arctan(2x−3)f(x)=arctan(2x-3)?