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

May 27, 2018

The domain is $x \in \left(\mathbb{R} - 3\right)$
And range is $f \left(x\right) \in \left(5 , \infty\right)$

#### Explanation:

in the function $f \left(x\right) = \frac{1}{{\left(x - 3\right)}^{2}} + 5$
you can see that if we put value of $x = 3$ then the function becomes undefined as we get $\left[\frac{1}{0}\right]$.

Thus we can put any value other than $3$. Thus the domain of the function is $x \in \left(\mathbb{R} - 3\right)$.

Now, to find the range find the inverse of the function $f \left(x\right)$ which is ${f}^{-} 1 \left(x\right)$.

let is consider $f \left(x\right)$ as $y$. So we can write--

$y = \frac{1}{{\left(x - 3\right)}^{2}} + 5$
rArr y-5 = 1/((x-3)^2
$\Rightarrow \frac{1}{y - 5} = {\left(x - 3\right)}^{2}$
$\Rightarrow \pm \frac{1}{\sqrt{y - 5}} = x - 3$
$\Rightarrow 3 \pm \frac{1}{\sqrt{y - 5}} = x$

Now for the function $\left\{\sqrt{y - 5}\right\}$ to be real we must have $y - 5 \ge 0$

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

$y - 5 > 0$

$\Rightarrow y > 5$

As $f \left(x\right) = y$

we get $f \left(x\right) > 5$
Hence the Range of the function is $\left(5 , \infty\right)$.