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

fattydq · #7 December 3rd, 2008, 02:06 PM

Quote:

Originally Posted by running-gag

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

You seem to be misunderstanding. I don't need to find where the function is increasing or decreasing, I just have to find the absolute min and absolute max values of the function. That's it. Because it's absolute min/max and not just relative min/max, I plugged in the roots -1 and 0, which gave me 1, which I tried for both the min and max answer, and it turned out it is indeed the min. Now when I solve 3x^2-1=0, the resulting fraction that you provided, and I already tried, gives a number LOWER than the min when plugged into the function so it CANNOT be the max. That's precisely why I'm confused about this particular problem.