How do you find the domain and range of # f(x)=1/(x-3)^2+5#?

1 Answer
May 27, 2018

Answer:

The domain is #x in (RR-3)#
And range is #f(x)in (5,oo)#

Explanation:

in the function #f(x)= 1/((x-3)^2) +5#
you can see that if we put value of #x=3# then the function becomes undefined as we get #[1/0]#.

Thus we can put any value other than #3#. Thus the domain of the function is #x in (RR-3)#.

Now, to find the range find the inverse of the function #f(x)# which is #f^-1(x)#.

let is consider #f(x)# as #y#. So we can write--

#y = 1/((x-3)^2) +5#
#rArr y-5 = 1/((x-3)^2#
#rArr 1/(y-5) = (x-3)^2#
#rArr +-1/sqrt(y-5) = x-3#
#rArr 3 +- 1/sqrt(y-5) = x#

Now for the function #{sqrt(y-5)}# to be real we must have #y-5 >= 0#

But since #y-5# is in denominator we have to consider another case which will give us

#y-5 > 0#

#rArr y >5#

As #f(x)=y#

we get #f(x) > 5#
Hence the Range of the function is #(5,oo)#.