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

Jul 15, 2018

See graphs and explanation.

#### Explanation:

$y = \pi \arcsin \left(4 x\right)$. Inversely, $x = \left(\frac{1}{4}\right) \sin \left(\frac{y}{\pi}\right)$.

The y-axis is the axis of this wave.

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

Amplitude =1/4 $\Rightarrow \left\mid x \right\mid \le 0.25 .$.

The range : $y \in \left[- {\left(\pi\right)}^{2} / 2 , {\left(\pi\right)}^{2} / 2\right] = \left[- 4.935 , 4.935\right]$.

Domain ; $x \in \left[- \frac{1}{4} , \frac{1}{4}\right]$.

See the graph of $y = \pi \arcsin \left(4 x\right)$, within the above guarding

limits, attributed to the convention

$\arcsin \left(4 x\right) \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$.
graph{y-3.14 arcsin (4x ) = 0}

Now see the graph for $y = \pi {\left(\sin\right)}^{- 1} \left(4 x\right)$, using the

common-to-both inverse $x = \left(\frac{1}{4}\right) \sin \left(\frac{y}{\pi}\right)$. Here, the inverse is

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

pixel\march to $\pm \infty$. This is not possible in the other graph

above, for .$y = \pi \arcsin \left(4 x\right)$..

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 ${\left(\sin\right)}^{- 1}$ operator and display display

${\left(\sin\right)}^{- 1} \left(\sin \left(- {120}^{o}\right)\right)$ as $- {120}^{o}$ and not $- {60}^{o}$, ..

.