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

The function will be concave in $\left(1 , \setminus \infty\right)$

The function will be convex in $\left(- \setminus \infty , 1\right)$

#### Explanation:

Given function:

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

$f ' \left(x\right) = \setminus \frac{\left(x - 1\right) \left(- 4 x\right) - \left(- 2 {x}^{2}\right) \left(1\right)}{{\left(x - 1\right)}^{2}}$

$= \setminus \frac{- 2 {x}^{2} + 4 x}{{\left(x - 1\right)}^{2}}$

$f ' ' \left(x\right) = \setminus \frac{{\left(x - 1\right)}^{2} \left(- 4 x + 4\right) - \left(- 2 {x}^{2} + 4 x\right) \left(2 \left(x - 1\right)\right)}{{\left(x - 1\right)}^{4}}$

$f ' ' \left(x\right) = - \frac{4}{x - 1} ^ 3$

The function will be concave iff $f ' ' \left(x\right) < 0$

$\setminus \therefore - \frac{4}{x - 1} ^ 3 < 0$

$\frac{1}{x - 1} ^ 3 > 0$

$x \setminus \in \left(1 , \setminus \infty\right)$

The function will be concave in $\left(1 , \setminus \infty\right)$

The function will be convex in $\left(- \setminus \infty , 1\right)$