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

1 Answer
May 27, 2018

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).