# Is f(x)=e^(2x)-x^2/(x-1)-1 concave or convex at x=0?

Jan 9, 2017

f(x) is convex at x=0

#### Explanation:

Let's calculate the first and the second derivative of f(x):

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

$f ' ' \left(x\right) = 4 {e}^{2 x} - \frac{\left(2 x - 2\right) {\left(x - 1\right)}^{2} - 2 \left(x - 1\right) \left({x}^{2} - 2 x\right)}{x - 1} ^ 4$
$= 4 {e}^{2 x} - \frac{2 {\left(x - 1\right)}^{{\cancel{3}}^{2}} - 2 \cancel{\left(x - 1\right)} \left({x}^{2} - 2 x\right)}{x - 1} ^ \left({\cancel{4}}^{3}\right)$
$= 4 {e}^{2 x} - \frac{2 {\left(x - 1\right)}^{2} - 2 \left({x}^{2} - 2 x\right)}{x - 1} ^ 3$

Then let's look at the sign of f''(0):

$f ' ' \left(0\right) = 4 {e}^{0} - \frac{2}{-} 1 = 4 + 2 = 6 > 0$

Then f(x) is convex at x=0

graph{e^(2x)-x^2/(x-1)-1 [-5.164, 5.17, -2.583, 2.58]}