How to find the range of a function?

**^_^Engineer_Adam^_^** · November 9th 2006, 02:09 AM

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

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

Ive got another question:
Why is $x = y^2$ not a function and $x = y^3$ is?

**ThePerfectHacker** · November 9th 2006, 09:05 AM

Originally Posted by ^_^Engineer_Adam^_^

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

cud u please explain how its done?

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

But the original function,
$y=\sqrt{4-x^2}$
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.

Why is $x = y^2$ not a function and $x = y^3$ is?

The curve $x=y^2$ cannot be a function. Because for example, $(-4,2),(-4,-2)$ 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 $x=y^3$ the only solution (real) for y is $y=\sqrt[3]{x}$ .

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

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