SOLUTION: Hi. Can I have some help finding the domain of this function? g(x)=√x/2x^2+x-1

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Question 125480: Hi. Can I have some help finding the domain of this function?
g(x)=√x/2x^2+x-1
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
I think you meant the function to be g%28x%29=sqrt%28x%29%2F2x%5E2%2Bx-1 rather than g%28x%29=sqrt%28x%2F2x%5E2%2Bx-1%29

The domain of a function is the set of all real values that x can take on such that the function is defined. I'm also presuming that you are mapping the function to R%5E2.

Since sqrt%28x%29 is undefined if x%3C0, and sqrt%28x%29%2F2x%5E2 is undefined at zero, the domain range is (0,infinity)

If, on the other hand you really meant the alternative function above with the entire expression under the radical, you could immediately say that zero is excluded from the domain, but you would have to work out the value for x that would cause x%2F2x%5E2%2Bx-1 to be less than zero.

x%2F2x%5E2%2Bx-1%3C0

Multiply by 2x%5E2
x%2B2x%5E3-2x%5E2%3C0
2x%5E3-2x%5E2%2Bx%3C0

Now, if we make a graph of f%28x%29=2x%5E3-2x%5E2%2Bx, we should be able to see where the function is less than zero, at least in an approximate sense.

graph%28600%2C600%2C-3%2C3%2C-3%2C3%2C2x%5E3-2x%5E2%2Bx%29

Looking at this graph, we see that the only real number zero of the function is at zero, so the interval for which f%28x%29=2x%5E3-2x%5E2%2Bx%3C0 is where x%3C0.

That means that the domain of g%28x%29 is the same regardless of whether g%28x%29=sqrt%28x%29%2F2x%5E2%2Bx-1 or g%28x%29=sqrt%28x%2F2x%5E2%2Bx-1%29, namely: (0,infinity)


The following graphs are of both original functions. The green one is the first way, and the red one is the one with the entire expression under the radical. You can see that the graph visually supports the analysis of the domain interval.