Math Help Forum - View Single Post - Absolute minimum and maximum of a function involving e

running-gag · #6 December 3rd, 2008, 01:58 PM

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

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

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

On [-1,0] you can find only one value for which $3x^2 - 1 = 0$
This value is $- \frac{\sqrt{3}}{3}$

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