# How do you find the maximum, minimum and inflection points for each function (x)/(x^2-9) ?

Jul 23, 2018

No relative maximum or minimum
Inflection at $\left(0 , 0\right)$

#### Explanation:

Given: $f \left(x\right) = \frac{x}{{x}^{2} - 9} = \frac{x}{\left(x - 3\right) \left(x + 3\right)}$

Find critical values: $f ' \left(x\right) = 0$:

Find the first derivative using the quotient rule:

$\left(\frac{u}{v}\right) ' = \frac{v u ' - u v '}{v} ^ 2$

Let u = x; " "u' = 1

Let v = x^2-9; " "v' = 2x

$f ' \left(x\right) = \frac{\left({x}^{2} - 9\right) \left(1\right) - x \left(2 x\right)}{{x}^{2} - 9} ^ 2 = \frac{{x}^{2} - 2 {x}^{2} - 9}{{x}^{2} - 9} ^ 2 = \frac{- {x}^{2} - 9}{{x}^{2} - 9} ^ 2$

$f ' \left(x\right) = \frac{- \left({x}^{2} + 9\right)}{{x}^{2} - 9} ^ 2 = 0$

-(x^2 +9) = 0; " "x^2 + 9 = 0; " "=> x = +-3i

No critical values.

Inflection points are found when $f ' ' \left(x\right) = 0$

Find the second derivative:

Let u = -x^2 - 9; " "u' = -2x

Let v = (x^2 - 9)^2; " "v' = 2(x^2-9)(2x) = 4x(x^2-9)

Find $f ' ' \left(x\right) = \frac{\left({\left({x}^{2} - 9\right)}^{2}\right) \left(- 2 x\right) - \left(- {x}^{2} - 9\right) \left(4 x \left({x}^{2} - 9\right)\right)}{{\left({x}^{2} - 9\right)}^{4}}$

$f ' ' \left(x\right) = \frac{\left({\left({x}^{2} - 9\right)}^{2}\right) \left(- 2 x\right) + \left({x}^{2} + 9\right) \left(4 x \left({x}^{2} - 9\right)\right)}{{\left({x}^{2} - 9\right)}^{4}}$

$f ' ' \left(x\right) = \frac{\left({x}^{2} - 9\right) \left[- 2 x \left({x}^{2} - 9\right) + 4 x \left({x}^{2} + 9\right)\right]}{{\left({x}^{2} - 9\right)}^{4}}$

$f ' ' \left(x\right) = \frac{- 2 {x}^{3} + 18 x + 4 {x}^{3} + 36 x}{{x}^{2} - 9} ^ 3 = \frac{2 {x}^{3} + 54 x}{{x}^{2} - 9} ^ 3$

$f ' ' \left(x\right) = \frac{2 x \left({x}^{2} + 27\right)}{{x}^{2} - 9} ^ 3 = 0$

$2 x \left({x}^{2} + 27\right) = 0$

$x = 0 , x = \pm \sqrt{27} i = \pm 3 \sqrt{3} i$

Inflection point at $x = 0$

$f \left(0\right) = 0$

Inflection at $\left(0 , 0\right)$