Let f: R - \ 2, 6\ → R be real valued function defined as f(x) = x...

Given,

$f: R - \left\{2, 6\right\} \to R$ be real valued function defined as $f\left(x\right) = \dfrac{x + 2x + 1}{{x}^{2} - 8x + 12}$

Now let, $y = \dfrac{{x}^{2} + 2x + 1}{{x}^{2} - 8x + 12} = \dfrac{{\left(x + 1\right)}^{2}}{\left(x - 2\right)\left(x - 6\right)} \ldots \left(1\right)$

Now differentiating the above function we get,

$⇒ \dfrac{dy}{dx} = \dfrac{- 2\left(x + 1\right)\left(5x - 16\right)}{{\left(x - 2\right)}^{2}{\left(x - 6\right)}^{2}}$

Now by wavy curve method we get,

So Graph of y = \dfrac{{\left(x + 1\right)}^{2}}{\left(x - 2\right)\left(x - 6\right)} for given domain will be,

So, from graph range is $y \in \left(- \infty , \dfrac{21}{4}\right] \cup [ 0, \infty )$