How do you sketch the graph of #f(x)=piarcsin(4x)#?

Question

2200 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-15T12:47:39

See graphs and explanation.

#y = pi arcsin (4 x )#. Inversely, #x = (1/4) sin ( y/pi )#.

The y-axis is the axis of this wave.

The wave-period = #(2pi)/(1/pi) = 2(pi)^2 = 19.74, nearly.

Amplitude =1/4 #rArr abs x <= 0.25.#.

The range : #y in [-(pi)^2/2, (pi)^2/2 ] = [ - 4.935, 4.935 ]#.

Domain ; #x in [ -1/4, 1/4 ]#.

See the graph of #y = pi arcsin (4 x )#, within the above guarding

limits, attributed to the convention

#arcsin ( 4 x ) in [ - pi/2, pi/2 ]#.
graph{y-3.14 arcsin (4x ) = 0}

Now see the graph for #y = pi (sin )^( -1 )( 4x )#, using the

common-to-both inverse #x = (1/4) sin (y/pi )#. Here, the inverse is

wholesome and, by sliding the graph #uarr datt #, you can see the

pixel\march to #+- oo#. This is not possible in the other graph

above, for .#y = pi arcsin (4 x )#..

graph{x-0.25 sin( y / 3.14)=0[ -1 1 -40 40] }

I expect that the hand calculators would soon use the

piecewise-wholesome #(sin)^(-1 ) # operator and display display

#(sin)^(-1)(sin (-120^o))# as #-120^o# and not #- 60^o#, ..

.