Derivative of exponential function

[domyy]

Homework Statement

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

f(x) = x²e^-x

Homework Equations

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?

 

[Dick]

Homework Statement

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

f(x) = x²e^-x

Homework Equations

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. Meaning, if I missed something, 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 to find extrema. Meaning, will 0 ALWAYS be part of the critical point and be used to find the MIN/MAX extrema?

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.

 

[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?

 

[Dick]

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?

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

 

[domyy]

Thanks!

 

[domyy]

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

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?

 

[Dick]

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?

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

 

[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?

 

[Dick]

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

Until now, am I correct?

That looks ok so far.

 

[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?

 

[domyy]

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

 

[Dick]

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?

It agrees with what I got.

 

[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.

 

[Dick]

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.

Use the quadratic equation. Irrationals are numbers too.

 

[domyy]

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

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

 

[Dick]

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

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

Complete the square or use the quadratic equation. http://en.wikipedia.org/wiki/Quadratic_equation

 

[Mark44]

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

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[/color](2xe^-x -x²e^-x)

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

Until now, am I correct?

 

[Dick]

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[/color](2xe^-x -x²e^-x)

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.