1. ## domain & range

what is the domain, range for these equations? kindly include explaination.

1. z = 4x² + y²
2. z = √(x² + y² - 16)
3. z = √(4 - y²)
4. z = 1/x²y²

2. Originally Posted by fuzzyorama
what is the domain, range for these equations? kindly include explaination.

1. z = 4x² + y²
2. z = √(x² + y² - 16)

1)The domain is all RxR. The range is >=0 because x^2>=0 and y^2>=0.

2)We require that x^2+y^2>=16. And the range is >=0 because sqrt( ) function can get arbitrarily big.

3. Hello, fuzzyorama!

What is the domain, range for these equations?

$\displaystyle 1)\; z\:=\:4x^2 + y^2$
$\displaystyle x$ can be any real value.
$\displaystyle y$ can be any real value.
. . Domain: .$\displaystyle x,y \in (-\infty,\,\infty)$

Since $\displaystyle x^2$ and $\displaystyle y^2$ are nonnegative, $\displaystyle z$ will be nonnegative.
. . Range: .$\displaystyle z \,\geq\,0$

$\displaystyle 2)\; z \:= \:\sqrt{x^2 + y^2 - 16}$

. . Domain: .$\displaystyle x^2+y^2\,\geq\,16$

$\displaystyle z$ will be nonnegative.
. . Range: .$\displaystyle z \,\geq\,0$

$\displaystyle 3)\; z \:= \:\sqrt{4 - y^2}$

The radicand must be nonegative. .$\displaystyle 4-y^2\:\geq\:0$
. . Domain: .$\displaystyle -2\:\leq \:y\:\leq\:2$

$\displaystyle z$ will be nonnegative.
. . Range: .$\displaystyle z\,\geq\,0$

$\displaystyle 4)\; z \:= \:\frac{1}{x^2y^2}$

$\displaystyle x,\,y$ cannot be zero.
. . Domain: .$\displaystyle x,y\:\neq\:0$

Since $\displaystyle x^2y^2$ is always positive, $\displaystyle z$ is positive.
. . Range: .$\displaystyle z\,>\,0$