Question

Range of this?

Answer 1

\[\large \sin^{-1} (\sqrt{x^2+x+1})\]

Answer 2

Attempt:
For the parameter of arcssin function..
I found the max value..it turned out to be infinity
and minimum value to be 1

Answer 3

\[1<\sqrt{x^2+x+1}<=\infty\]
\[\frac{\pi}{2}<\sin^{-1}\sqrt{x^2+x+1}<=?\]
how can we take sin inverse of infinity?

Answer 4

@hartnn @yrelhan4

Answer 5

@agent0smith

Answer 6

Isn't the range of arcsin limited between -1 and 1? It's defined that way or else it would not be a function (since sinx is not one to one)

Answer 7

Yeah i just plotted and google and saw that it's not in this case.

Answer 8

@agent0smith yes but the parameter of arcssin can have value from 0 and infinity

Answer 9

So it would be
\[\frac{\pi }{3}\leq \sin^{-1} (\sqrt{x^2+x+1}) \leq \frac{\pi }{2}\]

Answer 10

What do you think?

Answer 11

answer is correct but where did I go wrong with my approach?

Answer 12

Isn't the domain -1<x<0

Answer 13

\[-1 \leq x \leq0\]

Answer 14

I think you forgot to account for the fact that arcsin has a domain of -1 to 1, since sinx has a range of -1 to 1.

Answer 15

Yeah

Answer 16

and in this case\[\large \sqrt{x^2+x+1} \] can never be less than 0, and you can;t take the arcsin of it when it's greater than 1, so... domain -1 to 0 like @.Sam. said

Answer 17

\[\sin^{-1}(1)=90\\ \\ \sin^{-1}(2)= \infty\]

Answer 18

hmm, @.Sam. according to wolfram it's a complex number: http://www.wolframalpha.com/input/?i=arcsin2

it shouldn't be infinity, because arcsin is the inverse of sinx, range of sinx is -1 to 1, which becomes the domain of arcsinx, by the property of inverse functions.

Answer 19

For simplicity :P

Answer 20

ah :P I have never studied arcsins of numbers>|1|... wasn't aware they were complex numbers

Answer 21

i think its done, so this is for fun :)
\[x^2+x+1=(x+\frac{1}{2})^2+\frac{3}{4}\ge \frac{3}{4} \]\[\sqrt{x^2+x+1} \ge \frac{\sqrt{3}}{2}\]arcsin takes this part only\[\frac{\sqrt{3}}{2} \le \sqrt{x^2+x+1} \le 1\]so\[\frac{\pi}{3} \le \arcsin\sqrt{x^2+x+1} \le \frac{\pi}{2}\]

Answer 22

^nice