The domain of g is easy. g is a polynomial function and any number can be an input into a polynomial. So the domain is all real numbers.
The range is not so easy. Since g has an even degree, 4, and since the leading coefficient is positive, we know that the graph will ascend on both the left and right. So the range has no upper limit. There is a lower limit to the range.
If your calculator can do graphs and if it has a trace function, then you can find the minimum that way.
If not, then the only way I can think of to find the minimum is to use Calculus. If you do not know anything about derivatives then the following will not make sense. If you want an alternate explanation for how to find the minimum value for g, then you'll have to re-post your question.
Finding minimums using derivatives. First we find the first derivative:
g'(x) =
The relative maximums and minimums will be where this derivative is zero so we set it equal to zero and solve:
To solve this we factor it. 1 is a possible rational root and, as it turns out, it is actually a root. So x-1 is a factor and we can find the other factor using synthetic division:
1 | 4 -9 4 1
4 -5 -1
----------------
4 -5 -1 0
The zero in the lower right corner is the remainder. It is also the value of g' when x = 1. So we can see that 1 is actually a root of g'. This makes (x-1) a factor. And the rest of the bottom row tells us what the other factor is. The "4 -5 -1" translates into . Rewriting our equation as we've factored it so far:
The second factor will not factor further. But it is a quadratic so we can use the Quadratic Formula to find its roots. With the formula you should find that the roots of the quadratic factor are: .
So there are three roots of g': 1, and . One of these three will be the x value that has the lowest (or minimum value) for g.
One could find g(1), , and to see which is lowest.
But we could also use the second derivative to help us narrow down the possibilities:
g''(x) =
Now we find the value of g'' for our three candidates. We are only interested in whether g'' works out positive, negative or zero.
It is easy to see that g''(1) = -2. The fact that g''(1) is negative tells us that the graph is concave downward at x = 1. Concave downward for a root of g' means there is a relative maximum.
If you use your calculator to find decimals for the other two roots of g' and then find the value of g'' for each of them, you will find that g'' is positive for each one. A positive g'' for a root of g' means the graph is concave upward and, therefore, a relative minimum. So one of these will give us the lowest/minimum value for g.
Last of all we find , and , using the decimals we used in g'', to see which one results in a lower value for g. You should find that gives you the lowest value, somewhere around -1.2.
This makes the range, finally, all numbers greater than approximately -1.2. (If you really want an exact minimum, do not use a decimal for . This will give you an exact value for the minimum but the calculations to find it will be much more difficult.
Below you will see a graph for g(x) and you can see the that there is no upper limit to the y's and that the lowest y is at about -1.2: