Absolute minimum and maximum of a function involving e

fattydq · #1 $Old$ December 3rd, 2008, 10:35 AM

I have no problem finding maximum and minimums of functions, except when there's an e involved. For example finding the max and min values of e^((x^3)-x) on the closed interval -1 and 0. First off I don't know how to find the derivative of a power raised to another power, and even when I cheat and use a derivative calculator to find that, I don't know how to set the remaining derivative with e equal to 0. Would someone mind helping me out here?

running-gag · #2 $Old$ 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 !

fattydq · #3 $Old$ December 3rd, 2008, 01:31 PM

But I can't...I still don't understand how to set that derivative equal to 0 since there's an e in the term.

running-gag · #4 $Old$ December 3rd, 2008, 01:39 PM

The derivative of $e^{x^3-x}$ is $(3x^2 - 1)e^{x^3-x}$
This is a product therefore to set the derivative equal to 0 means solving 3x²-1 = 0 (since e^anything can never be equal to 0)

fattydq · #5 $Old$ December 3rd, 2008, 01:45 PM

But that makes no sense because and absolute min I already found to be 1. For that to be equal to 0, x would have to be much smaller than one...

running-gag · #6 $Old$ 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

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

Quote:

Originally Posted by running-gag $View Post$

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.

running-gag · #8 $Old$ December 3rd, 2008, 02:26 PM

I can confirm you that the absolute min are given for x=-1 and x=0 and this value is 1
I can also confirm you that the absolute max is given for $x = - \frac{\sqrt{3}}{3}$ and this value is approximately 1.469

running-gag · #9 $Old$ December 3rd, 2008, 02:30 PM

Here is the curve

fattydq · #10 $Old$ December 3rd, 2008, 03:04 PM

Oh jeez. I kept plugging in only POSITIVE sqrt 3/3, rather than negative. My apologies. This leads me to another similar question, however. What would you do if you have a similar function with e involved, and are asked to find the same thing (absolute min/max values) but the exponent's derivative no longer contains an x.

F(x) = e^-x - e^-2x

When you pull the exponents out in front there are no longer x's attached so how would you set it equal to 0 if e to any power can never equal 0.

running-gag · #11 $Old$ December 4th, 2008, 10:08 AM

In the case $F(x) = e^{-x} - e^{-2x}$
the derivative is $F'(x) = -e^{-x} +2 e^{-2x}$

F'(x) = 0 leads to $-e^{-x} +2 e^{-2x} = 0$

$2 e^{-2x} = e^{-x}$

$2 = e^x$

x = ln(2) which is an absolute max