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

running-gag · #2 December 3rd, 2008, 10:47 AM

Hi

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

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

To find the sign you have to remember that $e^{anything}$ is always strictly positive

Therefore the sign of the derivative of $e^{x^3-x}$ is the same as the sign of $3x^2 - 1$

Now I think that you can finish the job !