How do you find the domain and range of #f(x) =sqrt( x - 1) - 1/sqrt( 2-x)#?

1 Answer
May 30, 2017

Answer:

#1 <= x< 2#
#f(x) <= f(1.318)#

Explanation:

We have #f:x|->sqrt(x-1)-1/sqrt(2-x)#

Whenever we have a function with a square root, we know that the square root has to be bigger than or equal to #0#.

#sqrt(x-1) >= 0#

#x-1 >= 0#

#x >= 1#

For the second square root, we need to be more careful, because we have #1"/"sqrt(2-x)#. This means that #sqrt(2-x) > 0#.

#2-x > 0#

# 2 > x#

So our domain is #1 <= x < 2#

To find the range, we need to find the stationary point and find if the function is increasing or decreasing.

#f'(x)=1/2(1/sqrt(x-1)-1/(2-x)^(3/2))#

Since #1"/"(2-x)^(3/2) > 1"/"sqrt(x-1)#, #f'(x) < 0# so the function is decreasing. This means the function will have a maximum point, and all other #y#-values will be below that point.

Let #f'(x)=0#

#1/2(1/sqrt(x-1)-1/(2-x)^(3/2))=0#

Solving for #x#, we get #x=1.318#.

Therefore, #f(x) <= f(1.318)#