What is the range of $f(x) = 1 + sqrt(9 - x^2)$ ?

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Answer 1 · 2017-05-10T20:12:01

$1<=f(x)<=4$

The values that $f(x)$ can take are dependent on the values for which $x$ is defined.

So, in order to find the range of $f(x)$ , we need to find its domain and take evaluate $f$ at these points.

$sqrt(9-x^2)$ is only defined for $|x| <=3$ . But since we're taking the square of $x$ , the smallest value it can take is $0$ and the largest $3$ .

$f(0) =4$

$f(3)=1$

Thus $f(x)$ is defined over $[1,4]$ .

What is the range of f(x) = 1 + sqrt(9 - x^2)f(x) = 1 + sqrt(9 - x^2)?