Originally Posted by

**running-gag** You are studying the function $\displaystyle e^{x^3-x}$ on interval [-1,0]

The derivative of $\displaystyle e^{x^3-x}$ is $\displaystyle (3x^2 - 1)e^{x^3-x}$

To find the max and min of the function you have to solve $\displaystyle (3x^2 - 1)e^{x^3-x} = 0$ on [-1,0]

This is equivalent to solve $\displaystyle 3x^2 - 1 = 0$ on [-1,0]

On [-1,0] you can find only one value for which $\displaystyle 3x^2 - 1 = 0$

This value is $\displaystyle - \frac{\sqrt{3}}{3}$

Now you have to find the sign of the derivative on [-1,0] in order to know where the function is increasing and decreasing