# Math Help - domain, range, and trig

1. ## domain, range, and trig

well anyways, I'm working on applying domain and range to trig functions. I had a sheet of about 10 questions and 2 bonus ones. I got one bonus, but not the other and need help with another one on the work sheet.

Find the domain and range of:

f(x) = 2sin^-1(3x+4)-1

g(x) = sqrt[tan^-1(x^2 - 2) - pi/4]

2. Originally Posted by Mr_Green
f(x) = 2sin^-1(3x+4)-1
Let us begin with the domain.
The domain of y=asin(x) is [-1,1].

Thus, for the function,
f(x)=2asin(3x+4)-1

We require that,
-1<=3x+4<=1
-5<=3x<=-3
-5/3<=x<=-1

Now the range.
To find the range we use my little trick.
"The range of y=f(x) are all y such that the equation y=f(x) has at least one solution for x in the domain".

We begin by writing,
y=2*asin(3x+4)-1
Thus,
y+1=2*asin(3x+4)
Thus,
(y+1)/2=asin(3x+4)

In order to have a solution i.e.
(1/3)sin((y+1)/2)-4=x

We require that,
-pi/2<=(y+1)/2<=pi/2
-pi-1<=y<=pi-1

3. Originally Posted by Mr_Green
g(x) = sqrt[tan^-1(x^2 - 2) - pi/4]
Begin with the domain.

We notice that atan(x) is always defined for any x.

However, since there is a square root we have,
atan(x^2-2)-pi/4>=0
atan(x^2-2)>=pi/4
Take tangent of both sides (since the function is increasing).
x^2-2>=1
x^2>=3
x>=sqrt(3) or x<=-sqrt(3)

Now the range.
Begin by writing,
y=sqrt(atan(x^2-2)-pi/4)
In order to have a solution we require that y>=0 (because otherwise the square root is always positive).
Thus, if y>=0 then,
y^2=atan(x^2-2)-pi/4
y^2+pi/4=atan(x^2-2)
To take the tangent of both sides, that is,
tan(y^2+pi/4)=x^2-2

We require that,
-pi/2<=y^2+pi/4<=pi/4
Solving,
-sqrt(pi)/2<=y<=sqrt(pi)/2
But we already know that y>=0
Thus,
0<=y<=sqrt(pi)/2

Thus,
tan(y^2+pi/4)+2=x^2
To have a solution in the domain we require that,
The left hand side is positive, which it is, because y>=0.
Thus, the range is,
[0,sqrt(pi)/2]