How do you find the domain and range of (x^2-x-12)^-4?

Feb 16, 2018

$D o m = \mathbb{R} \setminus \setminus \left\{3 , - 4\right\}$
$R a n = \left(0 , \infty\right)$

Explanation:

(I guess you are working on $\mathbb{R}$, if not please tell me...)

We can rewrite this as

$\setminus \frac{1}{{\left({x}^{2} - x - 12\right)}^{4}}$

Let $f \left(x\right) = {\left({x}^{2} - x - 12\right)}^{4}$ and $g \left(y\right) = \setminus \frac{1}{y}$

The domain of the composition of functions $g f$ is given by

$\setminus \left\{x \setminus \in D o m \left(f\right) : f \left(x\right) \setminus \in D o m \left(g\right) \setminus\right\}$

Since the domain of a polynomial is the whole $\mathbb{R}$ and the domain of $g$ is $\mathbb{R} \setminus \setminus \setminus \left\{0 \setminus\right\}$, we have that the domain of $g f$ is

$\left\{x \setminus \in \mathbb{R} : {x}^{2} - x - 12 \setminus \ne 0\right\}$

Since ${x}^{2} - x - 12 = 0$ if and only if $x = - 4$ or $3$, we have that

$D o m = \mathbb{R} \setminus \setminus \left\{3 , - 4\right\}$

Now for the range, again we have that it is $g \left(R a n \left(f\right) \setminus \cap D o m \left(g\right)\right)$ .
The range of a quadratic polynomial with two distinct roots is

$\left[a , \infty\right)$

for some $a < 0$ (the vertex of the parabola defined by this quadratic equation).

Since now $4$ is even, we have that

${\left[a , \infty\right)}^{4} = \left[0 , \infty\right)$

since an even power of a negative number is positive. So from that we see that

$R a n = \left(0 , \infty\right)$

since for all $x \in \left(0 , \infty\right)$ there is $y \in \left(0 , \infty\right)$ such that $x = \frac{1}{y}$, namely $x = \frac{1}{y}$, and for the rest there is none.