# How do you determine whether the function f(x) = x/(x^2 - 5) is concave up or concave down and its intervals?

##### 1 Answer
Oct 19, 2015

See the explanation.

#### Explanation:

We have to find 2nd derivative,

$f ' \left(x\right) = \frac{{x}^{2} - 5 - x \cdot 2 x}{{x}^{2} - 5} ^ 2 = \frac{- {x}^{2} - 5}{{x}^{2} - 5} ^ 2$

$f ' ' \left(x\right) = \frac{- 2 x {\left({x}^{2} - 5\right)}^{2} - \left(- {x}^{2} - 5\right) 2 \left({x}^{2} - 5\right) \cdot 2 x}{{x}^{2} - 5} ^ 4$

$f ' ' \left(x\right) = \frac{- 2 x \left({x}^{2} - 5\right) - \left(- {x}^{2} - 5\right) 2 \cdot 2 x}{{x}^{2} - 5} ^ 3$

$f ' ' \left(x\right) = \frac{- 2 {x}^{3} + 10 x + 4 {x}^{3} + 20 x}{{x}^{2} - 5} ^ 3$

$f ' ' \left(x\right) = \frac{2 {x}^{3} + 30 x}{{x}^{2} - 5} ^ 3 = \frac{2 x \left({x}^{2} + 15\right)}{{x}^{2} - 5} ^ 3$

$f ' ' \left(x\right) = 0 \iff 2 x = 0 \left({x}^{2} + 15 \ne 0 \forall x \in R\right)$

$f ' ' \left(x\right) = 0 \iff x = 0$

${\left({x}^{2} - 5\right)}^{3} > 0 \forall x \in \left(- \infty , - \sqrt{5}\right) \cup \left(\sqrt{5} , + \infty\right)$
${\left({x}^{2} - 5\right)}^{3} < 0 \forall x \in \left(- \sqrt{5} , \sqrt{5}\right)$

$f ' ' \left(x\right) > 0 \forall x \in \left(- \sqrt{5} , 0\right) \cup \left(\sqrt{5} , + \infty\right)$ function is concave up

$f ' ' \left(x\right) < 0 \forall x \in \left(- \infty , - \sqrt{5}\right) \cup \left(0 , \sqrt{5}\right)$ function is concave down