+9491 Assuming that f(x) = 1 / (3x7)
Any x value that makes the denominator zero is excluded from the domain.
Let's find any excluded x value.
3x7 = 0 Divide both sides by 3 .
x7 = 0 Take the 7th root of both sides.
x = 0
So the only excluded x value is 0 . So the domain is (-∞, 0) U (0, ∞)
No matter how big or small of an x we plug in, f(x) cannot be zero.
The range is (-∞, 0) U (0, ∞)
Here's a graph of the function.
+2446
Hello again! I see that you are having difficulty finishing the complete simplification of \(\frac{9x+2}{3x^2-2x-8}+\frac{7}{3x^2+x-4}\).
I am glad to see that you have factored both quadratics correctly. Rewriting the denominators into its factored form results in \(\frac{9x+2}{(3x+4)(x-2)}+\frac{7}{(3x+4)(x-1)}\). Just like the one I solved over here at https://web2.0calc.com/questions/subtracting-rational-expressions#r1, your goal at the moment is to create a common denominator. Unlike the one I linked to above, finding it is a little bit more difficult. In that example, I noted how you want the LCM (least common multiple); however, you really only need a common multiple. It does not have to be the least, but the computation is easier if you do.
So, what is a common factor of \((3x+4)(x-2)\) and \((3x+4)(x-1)\)? The common factor of 3x+4 is relevant, but it alone will not be enough to figure out a common factor. To figure out a common multiple, just multiply both denominators together. This will always give you a common multiple. It may not be the least one, however. Multiplying both the denominators together yields \((3x+4)^2(x-2)(x-1)\). There is no need to expand yet. Notice, however, that both denominators have a factor of \(3x+4\), so we can divide that from the multiple we have. This gives us the least common multiple of \((x-2)(x-1)\).
Now, let's convert each fraction individually. I'll start with \(\frac{9x+2}{(3x+4)(x-2)}\).
| \(\frac{9x+2}{(3x+4)(x-2)}*\frac{(x-2)(x-1)}{(x-2)(x-1)}\) | I'm not necessarily concerned about expanding just yet. For now, let's just worry about getting the common denominator. |
| \(\frac{(9x+2)(x-2)(x-1)}{(3x+4)(x-2)^2(x-1)}\) | |
Now, let's convert the next fraction \(\frac{7}{(3x+4)(x-1)}\):
| \(\frac{7}{(3x+4)(x-1)}*\frac{(x-2)(x-1)}{(x-2)(x-1)}\) | Let's do the multiplication! |
| \(\frac{7(x-2)(x-1)}{(3x+4)(x-2)(x-1)^2}\) | |
You might notice that there is an issue, though, because the expression \(\frac{(9x+2)(x-2)(x-1)}{(3x+4)(x-2)^2(x-1)}+\frac{7(x-2)(x-1)}{(3x+4)(x-2)(x-1)^2} \Rightarrow \frac{(9x+2)(x-1)}{(3x+4)(x-2)(x-1)}+\frac{7(x-2)}{(3x+4)(x-2)(x-1)}\). You'll see that I factored out some unnecessary factors to create that common denominator. Now, let's add those fractions together!
| \(\frac{(9x+2)(x-1)}{(3x+4)(x-2)(x-1)}+\frac{7(x-2)}{(3x+4)(x-2)(x-1)}\) | Now that the denominators are the same, we can combine the fractions. |
| \(\frac{(9x+2)(x-1)+7(x-2)}{(3x+4)(x-2)(x-1)}\) | Now, let's multiply the numerator. Find the product of the binomials and distribute the 7. |
| \(\frac{9x^2-7x-2+7x-14}{(3x+4)(x-2)(x-1)}\) | Let's combine those like terms! |
| \(\frac{9x^2-16}{(3x+4)(x-2)(x-1)}\) | This is NOT the final answer. This is not fully simplified because the numerator is a difference of squares. Be attentive! |
| \(\frac{(3x+4)(3x-4)}{(3x+4)(x-2)(x-1)}\) | There is a common factor in the numerator and denominator that cancel out. |
| \(\frac{3x-4}{(x-2)(x-1)}\) | Now, expand the denominator again. |
| \(\frac{3x-4}{x^2-3x+2}\) | This is fully simplified; there is nothing else to do here. |
Now, let's consider the restrictions. All the factors of the original problem \((3x+4),(x-2),\text{and}(x-1)\) cannot be divided by zero. Set them equal to zero to figure out the forbidden values.
\(\frac{3x-4}{x^2-3x+2},x\neq -\frac{4}{3},1,2\)
