Question

find the domain and range of (sin sqrt x)^-2

Answer 1

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

Answer 2

You could put it in your Ti Calculator in the y= area then press 2nd graph to get a table. It will give you the answer.

Answer 3

domain is
\[x\geq 0\] since you cannot take the square root of a negative number

Answer 4

plus you'll have a lot of holes because you're dividing by sin(sqrt(x)), so solve sin(sqrt(x)) = 0 to find which values of x to remove

Answer 5

range is
\[[1,\infty)\] since the biggest since can be is 1, and so the smallest
\[\frac{1}{\sin^2(\sqrt{x})}\] can be is 1

Answer 6

oh damn i totally forget about that sorry. yes there an infinite amount of places where the function is not defined

Answer 7

oh sry solve sin^2(sqrt(x)) = 0 to find the values to remove

Answer 8

any values for which
\[\sin(\sqrt{x})=0\]

Answer 9

like for example
\[\{0,\sqrt{\pi},\sqrt{2\pi},...\}\]