Question

f(x)= sin(arctanx). find the range of f.

Answer 1

So you have composed two functions,
\[h(x)= \sin(x) \ and \ g(x)=\arctan(x)\]
\[\rightarrow f=h\circ g \]
meaning
\[f(x)=h(g(x))\]
\[g:\mathbb{R} \rightarrow \left[ -1;1 \right]\]
\[h:\mathbb{R} \rightarrow [-\frac{\pi}{2};\frac{\pi}{2}]\]
And since
\[[-1;1] \in \mathbb{R} \rightarrow f\ is\ defined\ \forall x \in \mathbb{R}\]

And since arctan(x) is strictly increasing and continuous in [-1;1] ,
\[h(g(]-\infty;\infty[))=h([-1;1])=[\arctan(-1);\arctan(1)]\]
Meaning
\[f:\mathbb{R} \rightarrow [\arctan(-1);\arctan(1)]=[-\frac{\pi}{4};\frac{\pi}{4}]\]
so there's your domain

Answer 2

I mean, your range.... domain is
\[\mathbb{R}\]

Answer 3

(-1,1)

Answer 4

Nevermind... I did arctan(sin(x)) sorry... his answer is right because , using my previous function notation,
\[g:\mathbb{R} \rightarrow [-\frac{\pi}{2};\frac{\pi}{2}]\]
And
\[\sin([-\frac{\pi}{2};\frac{\pi}{2}])=[-1;1] \]
Because
\[-\frac{\pi}{4} \in [-\frac{\pi}{2};\frac{\pi}{2}]\ and\ \frac{\pi}{4} \in [-\frac{\pi}{2};\frac{\pi}{2}]\]
and
\[\sin(\pm \frac{\pi}{4})=\pm 1\]
which are the maxima attained by the sine function...

Answer 5

\[\arctan:\mathbb{R}\to\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\]

Answer 6

Yeah I kinda messed up on the beginning and defined g as arctan then did it as if g was sine... sorry

Answer 7

it okay i got it now!
thanks for your help!!

Answer 8

\[g(x)=[g(x)= \left| cosx-1 \right|. on [0,2\pi], what is the x value when g attains its maximum value?\]

Answer 9

if I'm reading it correctly the x value is \[\pi\]

Answer 10

g(x)=[g(x)=|cosx−1|.on[0,2π],what is the x value when g attains its maximum value?

Answer 11

hopefully this is more clear..

Answer 12

I'm sticking to my answer :)