Inverse Trig Question (1 Viewer)

rand_althor · Jun 4, 2015

Hey could someone help me with finding the range and domain of:

$y=x\arcsin{(x^2)}$

Also how would I graph the function?

Thanks!

leehuan · Jun 4, 2015

The x part doesn't affect the domain, since y=x exists for all real x.

$y=\arcsin { x } \quad has\quad domain\quad -1\le x\le 1\\ So\quad y=\arcsin { { x }^{ 2 } } \quad has\quad domain\quad -1\le { x }^{ 2 }\le 1\\ Since\quad { x }^{ 2 }\ge 0\quad for\quad all\quad real\quad x,\quad we\quad just\quad have\\ { x }^{ 2 }\le 1\\ Solve\quad to\quad get\quad -1\le x\le 1$

I think graphing that would be a bit out of the ordinary for 3U... I'd use the 4U method of multiplying ordinates.

integral95 · Jun 4, 2015

It's an odd function

well compared to the typical inverse sine function $y = sin^{-1}x$ your graph would be much lower than that in the domain

$0 \leq x \leq 1$

Once you get that part, use the property that it's an odd function to sketch the left hand side.

Cheefi · Jun 4, 2015

What is the range?

InteGrand · Jun 4, 2015

Cheefi said:
What is the range?

The range is $\left\{y\in \mathbb{R}:y\in \left[\frac{\pi}{2},\frac{\pi}{2} \right] \right\}$ .

Here's a useful sketch: http://www.graphsketch.com/?eqn1_co..._lines=1&line_width=4&image_w=850&image_h=525

Cheefi · Jun 4, 2015

Is it possible to find the range without sketching the function?

InteGrand · Jun 4, 2015

Cheefi said:
Is it possible to find the range without sketching the function?

Yes; the function $f:[-1,1]\to \mathbb{R},f(x)=x\sin^{-1} x^2$ is clearly monotonically increasing on [0, 1], as it is the product of two monotonically increasing non-negative functions ( $x$ and $\sin ^{-1} x^2$ ), and f is an odd function, so the extreme values of the function just occur at the endpoints of the domain.

Inverse Trig Question (1 Viewer)

rand_althor

Active Member

leehuan

Well-Known Member

integral95

Well-Known Member

Cheefi

New Member

InteGrand

Well-Known Member

Cheefi

New Member

InteGrand

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)