Finding the range of this equation problem.

Find the range of the function y=-3sin(3x)-2

I know that I can plug it in to my graphing calculator but I'm not sure like how exactly to tell what the range is =(.

My professor wants it in union/infinity form eg. (-Infinity,-2]U(4,Infinity) which makes it even more confusing to me.

Could someone please help?

Much appreciated.

Re: Finding the range of this equation problem.

Quote:

Originally Posted by

**mattyc33** Find the range of the function y=-3sin(3x)-2

$\displaystyle \begin{gathered} - 1 \leqslant \sin (t) \leqslant 1 \hfill \\ - 3 \leqslant - 3\sin (t) \leqslant 3 \hfill \\ - 5 \leqslant - 3\sin (t) - 2 \leqslant 1 \hfill \\ \end{gathered} $

Re: Finding the range of this equation problem.

Quote:

Originally Posted by

**Plato** $\displaystyle \begin{gathered} - 1 \leqslant \sin (t) \leqslant 1 \hfill \\ - 3 \leqslant - 3\sin (t) \leqslant 3 \hfill \\ - 5 \leqslant - 3\sin (t) - 2 \leqslant 1 \hfill \\ \end{gathered} $

Oh =) that is quite easy, suprised I didnt know that, I think what's confusing me is the notation that my prof. would like me to use

Would it just be something like [-5,-3sin(x)-2]U[-3sin(x)-2,1] ?

How to put a range in union formation.

How can I put:

http://latex.codecogs.com/png.latex?...\end{gathered}

Into union notation?

Would it be [-5,-3sin(t)-2]U[-3sin(t)-2,1] ?

Much appreciated.

Re: How to put a range in union formation.

Quote:

Originally Posted by

**mattyc33**

The range is $\displaystyle [-5,1]$.

There is no union about it.

Re: How to put a range in union formation.

Re: Finding the range of this equation problem.

Quote:

Originally Posted by

**mattyc33** Oh =) that is quite easy, suprised I didnt know that, I think what's confusing me is the notation that my prof. would like me to use

**Would it just be something like [-5,-3sin(x)-2]U[-3sin(x)-2,1] ?**

**no.**

$\displaystyle -5 \le y \le 1$

in interval notation, $\displaystyle [-5,1]$

Interval Notation

Re: How to put a range in union formation.

Quote:

Originally Posted by

**mattyc33**

Do you understand what the "range" of a function is? It **cannot** depend on the variable as this "answer" does.