# What are the inflections points of y= e^(2x) - e^x ?

By the Chain Rule, the first derivative is $y ' = 2 {e}^{2 x} - {e}^{x}$ and the second derivative is $y ' ' = 4 {e}^{2 x} - {e}^{x}$. Inflection points occur at the values of $x$ where the second derivative changes sign (from positive to negative or negative to positive).
Setting $y ' ' = 0$ leads to $4 {e}^{2 x} - {e}^{x} = 0$. The left-hand side of this equation can be factored as ${e}^{x} \left(4 {e}^{x} - 1\right) = 0$. Since ${e}^{x}$ is never zero, it follows that we just need to solve $4 {e}^{x} - 1 = 0$ to get $x = \ln \left(\frac{1}{4}\right) = - \ln \left(4\right) \setminus \approx - 1.386$.
You can check that $y ' ' = 4 {e}^{2 x} - {e}^{x}$ changes sign at this point by graphing it. Therefore, there is an inflection point at $x = - \ln \left(4\right)$. The second coordinate of this point can be found by plugging it into the original function to get ${e}^{2 \setminus \cdot \ln \left(\frac{1}{4}\right)} - {e}^{\ln \left(\frac{1}{4}\right)} = \frac{1}{16} - \frac{1}{4} = - \frac{3}{16}$. The inflection point is therefore $\left(- \ln \left(4\right) , - \frac{3}{16}\right)$.