# If f(x) = (x^2-4)/ (x-1), what is f(0), f(1/2), f(-2), f(x-2), f(r^2) and f(1/t)?

Dec 3, 2017

$f \left(0\right) = 4$
$f \left(\frac{1}{2}\right) = \frac{15}{2}$
$f \left(- 2\right) = 0$
$f \left(x - 2\right) = \frac{x \left(x - 4\right)}{x - 3}$
$f \left({r}^{2}\right) = \frac{{r}^{4} - 4}{{r}^{2} - 1}$
$f \left(\frac{1}{t}\right) = \frac{\left(\frac{1}{t} ^ 2\right) - 4}{\left(\frac{1}{t}\right) - 1}$

#### Explanation:

To find the expressions that replaces the x in f(x), we have to replace the x in the equation with whatever is in the bracket, so if

$f \left(x\right) = \frac{{x}^{2} - 4}{x - 1}$

$f \left(0\right) = \frac{{0}^{2} - 4}{0 - 1} = \frac{- 4}{- 1} = 4$

$f \left(\frac{1}{2}\right) = \frac{{\left(\frac{1}{2}\right)}^{2} - 4}{\left(\frac{1}{2}\right) - 1} = \frac{\left(\frac{1}{4}\right) - 4}{\left(\frac{1}{2}\right) - 1} = \frac{15}{2}$

$f \left(- 2\right) = \frac{{\left(- 2\right)}^{2} - 4}{\left(- 2\right) - 1} = \frac{0}{-} 3 = 0$

$f \left(x - 2\right) = \frac{{\left(x - 2\right)}^{2} - 4}{\left(x - 2\right) - 1} = \frac{{x}^{2} - 4 x + 4 - 4}{x - 3} = \frac{x \left(x - 4\right)}{x - 3}$

$f \left({r}^{2}\right) = \frac{{\left({r}^{2}\right)}^{2} - 4}{\left({r}^{2}\right) - 1} = \frac{{r}^{4} - 4}{{r}^{2} - 1}$

$f \left(\frac{1}{t}\right) = \frac{{\left(\frac{1}{t}\right)}^{2} - 4}{\left(\frac{1}{t}\right) - 1} = \frac{\left(\frac{1}{t} ^ 2\right) - 4}{\left(\frac{1}{t}\right) - 1}$