# Derivative of exponential function

1. Dec 18, 2012

### domyy

1. The problem statement, all variables and given/known data

Find extrema and points of inflection (then graph it).

f(x) = x2e-x

2. Relevant equations

3. The attempt at a solution

So, for f'(x) = xe-x(2-x)

Critical point(s):

f'(x) = 0

2-x = 0
x= 2

I have a question before continuing. Will my critical point include 0 besides 2?

I still have doubts about whether 0 is always part of the critical points because my prof., when working on a similar problem, stated that only the number found was the critical point and didn't use zero. He said that e-x cannot be equal to zero. Meaning he used 2 to find the extrema. What he did was using test numbers between 2 to see where it was increasing and decreasing (THAT if I didn't miss any step while copying it - I tend to rush a little when copying the notes on the board since he writes a lot).

However, with this problem, my book seems to use 0 as the other point to find extrema. For instance, when substituting 2 and 0 back into the original function, I'll have (2, 4e-2) and (0,0). Being the first , the MAX and the last, the MIN extrema.

Does that mean that 0 was the other critical point?
Meaning, will 0 ALWAYS be part of the critical point and be used to find the MIN/MAX extrema?

Last edited: Dec 18, 2012
2. Dec 18, 2012

### Dick

If f'(0)=0 (and it is here) then x=0 is a critical point. The equation you want to solve is x(2-x)=0. It has two roots. If your prof missed x=0 that was a mistake.

3. Dec 18, 2012

### domyy

So I'll have zero for x= 0 and (2-x)= 0.

What if I had e^-x(2-x)? Without the x? Then, would my critical point be only 2?

4. Dec 18, 2012

### Dick

Absolutely. 0 isn't ALWAYS a critical point. It only is when f'(0)=0.

5. Dec 18, 2012

### domyy

Thanks!

Last edited: Dec 18, 2012
6. Dec 18, 2012

### domyy

Just a curiosity: if I had only e-x(2-x), then only 2 would be my critical point. Then, how would I know whether the point found after substituting it back into the function is MAX or MIN, considering that you can't compare it with any other point?

7. Dec 18, 2012

### Dick

You can compare it with points that aren't critical. Like x=0,1,3 etc.

8. Dec 18, 2012

### domyy

Thank you for clarifying that for me. I was having doubts about it. So I'll continue the problem:

Since the problem is long, I will write the answers I found and will only solve here the step which I am having trouble with.

So, again:

f'(x) = xe-x(2-x)
f'(x) = 0
x= 0 and (2-x) = 0; x=2

Critical point(s): 2 and 0
Substituting it back into the original function, I will have:
(2,4e-2) and (0,0).

Extrema MAX: (2,4e-2)
Extrema MIN:(0,0)

Now, to find the point of inflection (the part I am having trouble with):

f''(x) = xe-x(2-x)
I figure that doing the distributive would be better (maybe that's where I went wrong?):

f''(x) = 2xe-x -x2e-x

f''(x) = [(2x)(e-x)' + (e-x)(2x)'] + [(-x2)(e-x)'+(e-x)(-x2)']

Until now, am I correct?

9. Dec 18, 2012

### Dick

That looks ok so far.

10. Dec 18, 2012

### domyy

f''(x) = [(2x)(e-x)' + (e-x)(2x)'] + [(-x2)(e-x)'+(e-x)(-x2)']

f''(x) = [(2x)(e-x)(-1) + (e-x)(2)] + [ (-x2)(e-x)(-1) + (e-x)(-2x)]

f''(x) = -2xe-x + 2e-x+x2e-x - 2xe-x

f''(x) = e-x(-2x + 2 + x2 - 2x)
f''(x) = e-x(-4x + 2 + x2)

How is it so far?

Last edited: Dec 18, 2012
11. Dec 18, 2012

### domyy

Thanks a lot! The help you, mentors, from this forum, provide is of tremendous relevance when I have doubts/can't understand something.

12. Dec 18, 2012

### Dick

It agrees with what I got.

13. Dec 18, 2012

### domyy

Now, I am having trouble with how to proceed from here.

f'(x)=0
x2-4x + 2=0

I was expecting to get a "proper" number.

14. Dec 18, 2012

### Dick

Use the quadratic equation. Irrationals are numbers too.

15. Dec 18, 2012

### domyy

I can't factor this out.
What should I do?

x(-4+x)=-2
x=-2/(-4+x) ?

16. Dec 18, 2012

### Dick

17. Dec 18, 2012

### Staff: Mentor

The two equations above are incorrect since you haven't found the 2nd derivative yet. What's on the right side is actually f'(x).

The corrected second equation would be
f''(x) = d/dx(2xe-x -x2e-x)

18. Dec 18, 2012

### Dick

Yeah, I'll usually gloss over bad notation on the preceding lines if the last one comes out correct. Good to correct that though for writing up the final problem.