- #1

SolCon

- 33

- 1

I have 2 simple questions, one is simply a confirming the answer query, and the other is domain related (like topic title says).

1) In this question, we are asked to show that a

^{2}+b

^{2}is constant for all values of [tex]\theta[/tex]. We have a and b as:

a=2sin[tex]\theta[/tex]+cos[tex]\theta[/tex]

b=2cos[tex]\theta[/tex]-sin[tex]\theta[/tex]

My final answer to this question was:

4sin

^{2}[tex]\theta[/tex]+sin

^{2}[tex]\theta[/tex]+4-4sin

^{2}[tex]\theta[/tex]+1-sin

^{2}[tex]\theta[/tex]

which results in simply a 5. Does this satisfy the question's demands?

2) In this question, we were asked to find an expression in terms of x, for f

^{-1}(x) and find the domain of f

^{-1}(x).

The original equation is: 6/(2x+3) for x>=0

So, for f

^{-1}(x), we'll have: x=6/(2y+3)

The final equation: y=1/2[(6/x)-3]

To get the domain, we place f

^{-1}(x)=0

So, 1/2[(6/x)-3]=0 , which will give us x=2, right?

After this, the domain must now be structured. But I'm having trouble understanding how this was done. The answer to the question is 0<x<=2. But why is it like this and not 0<=x<=2 instead?

Any help with this is appreciated.