Find the range of the function` f(x) = ((x^2 -4)(x-3))/(x^2-x-6).`

Thursday, April 14, 2011

Find the range of the function $\displaystyle{f{{\left({x}\right)}}}=\frac{{{\left({{x}}^{{2}}-{4}\right)}{\left({x}-{3}\right)}}}{{{{x}}^{{2}}-{x}-{6}}}.$

Hello!

The range of a function is the set of its possible values.

The given expression for the function $\displaystyle{f{}}$ is not so simple, let's try to simplify it. But first find the domain. There is a denominator, and it has not to be zero:

$\displaystyle{{x}}^{{2}}-{x}-{6}\ne{0},$ which gives $\displaystyle{x}\ne-{2}$ and $\displaystyle{x}\ne{3}$ (easy to guess).

Therefore $\displaystyle{{x}}^{{2}}-{x}-{6}={\left({x}+{2}\right)}{\left({x}-{3}\right)}.$ The numerator has something in common:

$\displaystyle{\left({{x}}^{{2}}-{4}\right)}{\left({x}-{3}\right)}={\left({x}-{2}\right)}{\left({x}+{2}\right)}{\left({x}-{3}\right)}.$

Thus the function becomes $\displaystyle{f{{\left({x}\right)}}}=\frac{{{\left({x}-{2}\right)}{\left({x}+{2}\right)}{\left({x}-{3}\right)}}}{{{\left({x}+{2}\right)}{\left({x}-{3}\right)}}}={x}-{2},$ a simple expression. But remember that $\displaystyle{x}\ne-{2}$ and $\displaystyle{x}\ne{3}.$

Now we can easily find the range. For $\displaystyle{x}-{2}$ when all x's are possible, it is the entire $\displaystyle\mathbb{R}.$ Because $\displaystyle{x}=-{2}$ and $\displaystyle{x}={3}$ are forbidden, the values $\displaystyle{y}=-{2}-{2}=-{4}$ and $\displaystyle{y}={3}-{2}={1}$ are impossible. So the range of $\displaystyle{f{}}$ is $RR\{-4,1},$ which is