How would you solve a problem Like this: Precalc!?

Question

For FYI in this question everything in { these } is in a square root sign. I didn't know how else to illustrate it.

I need to find the domain:

h(x)= 5{x/2x-1 - 3/x+2} + x^2

Any help on this would be appreciated, I have a Test soon >.<

I'm sorry! Let me rewrite the question with ( ) My teacher didn't use them . . . .

h(x)= 5 {(x/[2x-1]) - (3/[x+2])} + x^2

Answer 1

h(x)=
5[x(x+2)-3(2x-1)/(2x-1)(x+2)]+x^2
5[{x^2+2x-6x+3}/(2x^2-5x-2)]+x^2
[(5x^2-20x+15)+x^2(2x^2-5x-2)]
/(2x^2-5x-2)
that is as far aswe can go
we cannot solve as the RHS is absent

Answer 2

The domain of x^2 is all real x
The doman of the expression under the square root requires

x/(x+2) - 3/(x+2) >= 0

Solving this you will get (x-3)(x-1)>=0. This inequality is only satisfied for x<=1 or x>=3.

Combining the two (taking the union of the two domains), the domain of h is x<=1 or x>=3

Good luck on your test!

Answer 3

the domain is where the sqt is not negative

thus u should solve x/2x-1-3/x+2 < 0

since u dont use ( ) i do not know what you mean with x/2x-1 there fore i can not solve it for although i reallly want to....

Answer 4

h(x) can be graphed. Put it into your calculator. The domain is how far left and right on the x axis the graph goes. I think the domain of this graph is going to be infinity and negative infinity. Its hard to show you on the computer.