# What are the extrema of f(x) = (3x) / (x² - 1)?

Jan 2, 2016

The function contains no extrema.

#### Explanation:

Find $f ' \left(x\right)$ through the quotient rule.

$f ' \left(x\right) = \frac{\left({x}^{2} - 1\right) \frac{d}{\mathrm{dx}} \left(3 x\right) - 3 x \frac{d}{\mathrm{dx}} \left({x}^{2} - 1\right)}{{x}^{2} - 1} ^ 2$

$\implies \frac{3 \left({x}^{2} - 1\right) - 3 x \left(2 x\right)}{{x}^{2} - 1} ^ 2$

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

Find the turning points of the function. These occur when the derivative of the function equals $0$.

$f ' \left(x\right) = 0$ when the numerator equals $0$.

$- 3 \left({x}^{2} + 1\right) = 0$
${x}^{2} + 1 = 0$
${x}^{2} = - 1$

$f ' \left(x\right)$ is never equal to $0$.

Thus, the function has no extrema.

graph{(3x)/(x^2-1) [-25.66, 25.66, -12.83, 12.83]}