Question

define the implied domain and range of y=arctan(x^2)

We already know that the domain of arctan= R and range (-pi/2,pi/2)

Answer 1

For composite functions \(f(g(x))\), the range of \(f\) is restricted to the range of \(g\).

In this case, \(g(x)=x^2\), whose range is \(\{x~|~x\ge0\}\). For \(f(x)=\arctan x^2\), now that we're only considering non-negative values \(x^2\), we have a domain of \(\left\{f(x)~|~0\le f(x)\le\dfrac{\pi}{2}\right\}\).

Answer 2

Could you please explain it further because I really didnt get your explanation

Answer 3

i know that whatever the value of x is, y (range) will be greater than or equal to zero

Answer 4

correct me if im mistaken but isnt the domain: R and range: [0,pi/2)? I solved it by graphing y=arctan(x^2)

Answer 5

The domain and range are correct.

Like I said before, \(x^2\) will be non-negative. The arctangent of a positive number is always positive.

Answer 6

So if the arctangent is always going to be positive, you would reduce the range from \\(\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\) to \(\left[0,\dfrac{\pi}{2}\right)\).

Answer 7

and what will be the domain? shouldnt it be ALL REAL NUMBERS (R)

Answer 8

It is all real numbers, yes. \(x^2\) can take on any real number \(x\).

Answer 9

thanks :)

Answer 10

one more question. what if y=tan(x^2) how can we find the domain and range?

Answer 11

i know that the domain for y=tan(x) is R except pi/2+k*pi and the range: all real numbers

Answer 12

@SithsAndGiggles

Answer 13

i would really appreciate your help. Thanks

Answer 14

\[\tan x^2=\frac{\sin x^2}{\cos x^2}\]
\(\tan x^2\) is undefined when \(\cos x^2=0\), which occurs for \(x^2=\dfrac{\pi}{2}+k\pi\), or \(x=\pm\sqrt{\dfrac{\pi}{2}+k\pi}\) for positive integers \(k\) (negative integers have the potential to make the \(\cdots\) in \(\sqrt\cdots\) negative, and hence not real).

The range remains the same, all real numbers.