# How would you find the domain of each: y=(3x+2)/(4x+1)? y=(x^2-4)/(2x+4)? y=(x^2-5x-6)/(x^2+3x+18)? y=(2^(2-x))/(x)? y=sqrt(x-3)-sqrt(x+3)? y=sqrt(2x-9)/(2x+9)?

Nov 22, 2017

The domain would be all real numbers except for any value of $x$ that makes the statement impossible.

#### Explanation:

For example, we can look at your first statement, $\frac{3 x + 2}{4 x + 2}$.

Any value of $x$ that makes the denominator equal to 0 cannot be in the domain because you can't divide by zero.
To find what that value is, you can set up an equation like this:
$4 x + 2 = 0$

After solving this, you would get the answer of $x = - \frac{1}{2}$, which means that $x$ cannot equal $- \frac{1}{2}$, or the denominator will equal 0. So the domain for this one would be written $x : x \in \mathbb{R} , x \ne - \frac{1}{2}$

You would do the same for the other fraction problems. In the second problem, you would need to factor the denominator or use the quadratic formula to find the non-domain values.

For your square root problems, the domain would be any value of $x$ except for the ones that make the expression inside the sqrt negative, since you can't take the square root of a negative number.

For example, let's look at $y = \sqrt{x - 3} - \sqrt{x + 3}$
We can set up our equations again:
$x - 3 = 0$
$x + 3 = 0$

The answers here would be $x = 3$ and $x = - 3$.
This means that any value of $x$ that is lower than 3 (this would include -3, so we don't need to make any special considerations) will leave you with a value less that 0. None of these values can be part of the domain.
$x : x \in \mathbb{R} , x \ge 3$

The last question is an amalgam of these two problem types: $x$ can not equal anything that will make the denominator equal 0 or anything that will make the expression in the square root negative.

We can set up equations here again:
$2 x + 9 = 0$ --> $x = - \frac{9}{2}$
$2 x - 9 = 0$ --> $x = \frac{9}{2}$

So anything less than $\frac{9}{2}$ will make the square root negative, and $- \frac{9}{2}$ will make the denominator 0. Since $- \frac{9}{2} < \frac{9}{2}$, we don't need to add in an extra statement for it:
$x : x \in \mathbb{R} , x \ge \frac{9}{2}$

I hope that makes sense and helps you solve the fraction problems that I did not address.

Nov 22, 2017

#### Explanation:

Domain in $y = f \left(x\right)$ means thee values which $x$ can take. For finding domain we normally start in a reverseway i.e. values which $x$ cannot take.

For example if we have $x - a$ in denominator, as we cannot have denominator as $0$, we cannot have $x = a$. Similarly, if we have $\sqrt{x - a}$ as we cannot have square root og a negative number, we cannot have $x - a < 0$ or $x < a$.

In case we have a quadratic polynomial such as $a {x}^{2} + b x + c$ in denominator, we should either factorise it to $a \left(x - \alpha\right) \left(x - \beta\right)$ or convert it to form $a {\left(x - h\right)}^{2} + k$, to check restrictions on the value of $x$.

Sometimes factors may cancel out, if they are common between numerator and denominator. In that case, we call it a hole, because though $f \left(x\right)$ may be defined.

1. In $y = \frac{3 x + 2}{4 x + 1}$, domain isall $x$ other than $x = - \frac{1}{4}$, as latter makes denominator $0$.
2. In $y = \frac{{x}^{2} - 4}{2 x + 4} = \frac{\left(x + 2\right) \left(x - 2\right)}{2 \left(x + 2\right)} = \frac{x - 2}{2}$ and we cannot have $x = \pm 2$ and domain is values of $x$ other than $\pm 2$ and at $x = 2$, we have a hole.
3. As $y = \frac{{x}^{2} - 5 x - 6}{{x}^{2} + 3 x + 18} = \frac{\left(x - 6\right) \left(x + 1\right)}{{\left(x + \frac{3}{2}\right)}^{2} + \frac{63}{4}}$. Note that least value of denominator is $\frac{63}{4}$ and hence $y = f \left(x\right)$ exists for all vales of $x$ and hence domain is $\mathbb{R}$.
4. In $y = {2}^{2 - x} / x$, we have no restrictions on $x$ as far as numerator is concerned, however, denominator restricts domain of $x$ so that $x \ne 0$.
5. In $y = \sqrt{x - 3} - \sqrt{x + 3}$ , as we have $\sqrt{x - 3}$ we cannot have $x < 3$ and as we also have $\sqrt{x + 3}$, other restriction is we cannot have $x < - 3$. This means domain is $x \ge 3$.
6. As $y = \frac{\sqrt{2 x - 9}}{2 x + 9}$, numerator places a restriction that $x \ge \frac{9}{2}$ and denominator $x \ne - \frac{9}{2}$. This can be combined and domain is $x \ge \frac{9}{2}$.