# How do you find the relative extrema for f(x) = (x^2 - 3x - 4)/(x-2)?

May 15, 2017

That function has no relative extrema.

#### Explanation:

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

Domain of $f$ is $\left(- \infty , 2\right) \cup \left(2 , \infty\right)$

$f ' \left(x\right) = \frac{{x}^{2} - 4 x + 10}{x - 2} ^ 2$

$f ' \left(x\right)$ exists for all $x$ in the domain of $f$

$f ' \left(x\right) = 0$ where ${x}^{2} - 4 x + 10 = 0$. But this quadratic has discriminant $16 - 40$ which is negative. Therefore, there are no real solutions to the equation, and consequently there are no critical numbers for $f$.

Since relative extrema occur at critical numbers, there are no relative extrema for $f$.