Question

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

Answer 1

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)`.