# What is the domain and range of f(x) = -1 sqrt (1-x^2)?

Aug 3, 2015

Domain: $\left[- 1 , 1\right]$
Range: $\left[0 , - 1\right]$

#### Explanation:

The domain of the function will only be influenced by the fact that the square root of a number will result in a real value only if said number is positive.

In other words, the possible values of $x$ must satisfy the condition $\sqrt{1 - {x}^{2}} \ge 0$. First, find the zeros of this quadratic by

$\sqrt{1 - {x}^{2}} = 0$

$1 - {x}^{2} = 0$

$\left(1 - x\right) \left(1 + x\right) = 0$

This condition is met when $x = - 1$ and $x = 1$. Now, you need your quadratic to be greater than or equal to Zero, which means tha you need to find the interval(s) on which it is positive.

It's easy to see that for $x < - 1$ and $x > 1$, the quadratic will be negative, which means that the interval of interest will actually be $\left[- 1 , 1\right]$.

Since the only values of $x$ that will satisfy $\sqrt{1 - {x}^{2}}$ lie between $x = - 1$ and $x = 1$, the domain of the function will be $\left[- 1 , 1\right]$.

Now for the range of the function. Since the quadratic we've just looked at is equal to zero for $x = - 1$ and $x = 1$, the range of the function will vary between zero and the point in which the quadratic has a maximum value.

More specifically, $\sqrt{1 - {x}^{2}} = \text{max}$ if $x = 0$, since you're subtracting a squared value from $1$.

At point $x = 0$, the function is equal to

$f \left(0\right) = - \sqrt{1 - {0}^{2}} = - \sqrt{1} = - 1$

Therange of the function will thus be $\left[0 , - 1\right]$.

graph{-sqrt(1-x^2) [-10, 10, -5, 5]}