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

1 Answer
May 10, 2017

Answer:

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

Explanation:

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]#.