# For what values of x is f(x)= x^2-x + 1/x-1/x^2  concave or convex?

##### 1 Answer
Nov 19, 2017

$x = - 1 , 3 , \frac{3 \pm \sqrt{21}}{2}$

#### Explanation:

Find the second derivative of $f \left(x\right)$.
$f ' \left(x\right) = 2 x - 1 - \frac{1}{x} ^ 2 + \frac{2}{x} ^ 3$
$f ' ' \left(x\right) = 2 + \frac{2}{x} ^ 3 - \frac{6}{x} ^ 4$

Find the critical points of $f ' ' \left(x\right)$.
$0 = 2 + \frac{2}{x} ^ 3 - \frac{6}{x} ^ 4$
$0 = 2 \left(1 + \frac{1}{x} ^ 3 - \frac{3}{x} ^ 4\right)$
$- 1 = \frac{1}{x} ^ 3 - \frac{3}{x} ^ 4$
$- 1 = {x}^{3} - {x}^{4} / 3$
$- 1 = x \left({x}^{2} - {x}^{3} / 3\right)$

$x = - 1$

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

$x = 3$

${x}^{2} - 3 x = 3$
$0 = {x}^{2} - 3 x - 3$

Use the quadratic formula.

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{3 \pm \sqrt{21}}{2}$

$x = - 1 , 3 , \frac{3 \pm \sqrt{21}}{2}$