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

Jan 2, 2016

Study the sign of the 2nd derivative.

For $x < 1$ the function is concave.
For $x > 1$ the function is convex.

Explanation:

You need to study curvature by finding the 2nd derivative.

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

The 1st derivative:

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

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

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

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

The 2nd derivative:

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

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

$f ' ' \left(x\right) = 2 \cdot \left(- 2\right) {\left(x - 1\right)}^{-} 3$

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

Now the sign of $f ' ' \left(x\right)$ must be studied. The denominator is positive when:

$- {\left(x - 1\right)}^{3} > 0$
${\left(x - 1\right)}^{3} < 0$
${\left(x - 1\right)}^{3} < {0}^{3}$
$x - 1 < 0$
$x < 1$

For $x < 1$ the function is concave.
For $x > 1$ the function is convex.

Note: the point $x = 1$ was excluded because the function $f \left(x\right)$ can not be defined for $x = 1$, since the denumirator would become 0.

Here is a graph so you can see with your eyes:

graph{(-2x)/(x-1) [-14.08, 17.95, -7.36, 8.66]}