Question

If f(x) = √9-x^2, what are its domain and range ?

Answer 1

- The domain of a function f(x) is the set of all values for which the function is defined.

- The range of the function is the set of all values that f takes.

In our problem we have the following function:

`f(x)=\sqrt[]{9-x^2}`

Domain

In order to find the domain we must look for the values of x for which the function is defined.

We see that the function has a square root, if we want to have a function with real values, the argument of the square root must be positive. So we need to find the values of x such that:

`9-x^2\ge0`

From the last equation we see that:

`\begin{gathered} 0\leq9-x^2 \\ x^2\leq9 \\ -3\leq x\leq3 \end{gathered}`

So the values of x must me in the interval: [-3,3] (domain)

Why? Because if replace, for example x = 4, in the equation of our function, we see that we get the square root of a negative number!

Range

In order to find the range of the function, we must look for the values of f(x) in the domain of the function.

- First, we can see that we only can have positive values for our function. So in principle the domain is: [0,∞)

- Second, we must see if our function has a maximum value. We see that the maximum value of our function is get it when we replace x = 0, and we get: f(0) = √9 = 3

Taking in account the last results, we see that the range of our function is: [0,3]

Summary

- Domain: [-3,3]

- Range: [0,3]