# How to find the range of a function?

• Nov 9th 2006, 01:09 AM
How to find the range of a function?
Hello i got no problem with domain and im a bit of confused on how to find the range of a function given:
$\displaystyle f(x) = \sqrt{4-x^2}$
the answer is $\displaystyle [0,2]$

cud u please explain how its done? i really dont understand how to find the range

Ive got another question:
Why is $\displaystyle x = y^2$ not a function and $\displaystyle x = y^3$ is?
• Nov 9th 2006, 08:05 AM
ThePerfectHacker
Quote:

Hello i got no problem with domain and im a bit of confused on how to find the range of a function given:
$\displaystyle f(x) = \sqrt{4-x^2}$
the answer is $\displaystyle [0,2]$

cud u please explain how its done?

Here is a simple explaination.
$\displaystyle y=\sqrt{4-x^2}$
When you square it you have,
$\displaystyle y^2=4-x^2$
$\displaystyle x^2+y^2=2^2$
A circle!

But the original function,
$\displaystyle y=\sqrt{4-x^2}$
Is a semi-circle because we only take the positive side of the circle (since square root only returns positive values).

Since the radius it 2 it starts at zero and goes all the way until 2.

Quote:

Why is $\displaystyle x = y^2$ not a function and $\displaystyle x = y^3$ is?
The curve $\displaystyle x=y^2$ cannot be a function. Because for example, $\displaystyle (-4,2),(-4,-2)$ lie on this curve. Thus, for an absicca (x) values we have two ordinate (y) values which means it cannot be a function.

The other curve only has one value.
Because when you have $\displaystyle x=y^3$ the only solution (real) for y is $\displaystyle y=\sqrt[3]{x}$.

While in the first curve there are two solutions for y, $\displaystyle y=\pm \sqrt{x}$ (because it has even degree).