# Thread: [SOLVED] Natural Log Help

1. ## [SOLVED] Natural Log Help

I have to find the max. and min. values of the following function:

$y=3xe^{-x}+x$

When I take the first derivative, I get:

$y'=3e^{-x}-3xe^{-x}+1$

Though I'm not supposed to, I took a peak at the graph of the first derivative on my trusty graphing calculator and found that there are no critical values. So my question is, how do I show that there are no critical values using only calculation? (We are not allowed to use graphing calculators.)

(Sorry if I posted this in the wrong section. )

2. Hello,

The sign of y' is not obvious from a first sight.

So let's see what the second derivative is...

$y''=\dots=3e^{-x}(x-2)$

It is 0 if x=2. If x<2, y''<0, if x>2, y''>0

It means that y' decreases if x<2 and increases if x>2.
There is a minimum at x=2 for y'.

But $y'(2)=3e^{-2}-6e^{-2}+1=-\frac{3}{e^2}+1$

$e^2 \approx 2.6^2$ and is obviously >3.
So $\frac{3}{e^2}<1$

Hence $-\frac{3}{e^2}+1=m>0$
m is a minimum for y.

From this, you can deduce that y' is always positive, since for any x, we have $y' \geq m>0$.

Therefore, y is always increasing.

Now, study the limits of y as x goes to -infinity and +infinity, and see if you can have y=0