SOLUTION: Find the domain of h(x) = 3x^2 + 5x - 3. I got the answer (-2, -5). I don't believe that is correct, but it is what I got. Also, the domain of 3/x^2 + 7. I got (infinity, 15).

Question 142636: Find the domain of h(x) = 3x^2 + 5x - 3. I got the answer (-2, -5). I don't believe that is correct, but it is what I got. Also, the domain of 3/x^2 + 7. I got (infinity, 15). I am almost positive that is wrong, but the graph that showed up was very confusing for me.
Answer by nabla(475) (Show Source): You can put this solution on YOUR website!
The domain just means "for what values of x is the function defined?"
h(x)=3x^2 + 5x - 3
Can you find ANY number you can use that will not produce a real, defined number? I can't. Thus, the domain is all real numbers (double-strike R or (- infinity, infinity) in interval notation).
Now, g(x)=3/(x^2 + 7); For what values of x is this not defined? Only when the denominator is zero is when it cannot be defined. Thus, we would need x^2 to be negative. Can this ever be the case? In other words, can you think of any squared number that is REAL and negative? I can't. Thus, the domain is all real numbers (double-strike R or (- infinity, infinity) in interval notation).
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