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Family of functions etc

  • Thread starter t_n_p
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  • #1
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Homework Statement
http://img155.imageshack.us/img155/1432/img0193abz8.jpg [Broken]

The attempt at a solution

http://img256.imageshack.us/img256/8302/img0194aeb9.jpg [Broken]
I want to find the sign (i.e. max or min) of the function, however I'm not quite sure if my last step (deriv>0 for x>a/2) and hence deriv<0 for for x<a/2 is correct. I think this is a better idea than finding the double derivative because it starts to get messy.

Homework Statement





Homework Equations





The Attempt at a Solution

 
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Answers and Replies

  • #2
mjsd
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NB: x=0 is also a critical point
look at the 2nd derivative to check the nature of turning point.. and it doesn't get that messy... product rule is a bit like binomial theorem
 
  • #3
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NB: x=0 is also a critical point
look at the 2nd derivative to check the nature of turning point.. and it doesn't get that messy... product rule is a bit like binomial theorem
Oh yeh, you get the x=0 by dividing 0 by that exponential correct! :approve:

Using double derivative method I get...

http://img134.imageshack.us/img134/521/asdffz1.jpg [Broken]

But for f''(a/2) the answer will be in terms of a. Hope you can read what I've written, but how would I determine if f''(x) is positive? (I know a>0, but I'm still confused)
 
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  • #4
Dick
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Also the nonzero critical point is not at a/2. It's at 2/a. You might find it's a lot easier to recognize the sign of f'' if you put in the right number for the critical point.
 
  • #5
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Also the nonzero critical point is not at a/2. It's at 2/a. You might find it's a lot easier to recognize the sign of f'' if you put in the right number for the critical point.
silly me....:yuck:
 
  • #6
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I figure out f"(2/a) = -2e^(-2)
which is a negative number hence at x=2/a a maximum occurs

*thumbsup*

For part c) I'm assuming I just solve for f" = 0?
 
  • #7
HallsofIvy
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By the way, you very first statement, that "the limit, as |x| goes to infinity is 0, because e-x is 0 is wrong. x2e-x goes to 0 as x goes to + infinity, but increases without bound as x goes to negative infinity.
 
  • #8
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By the way, you very first statement, that "the limit, as |x| goes to infinity is 0, because e-x is 0 is wrong. x2e-x goes to 0 as x goes to + infinity, but increases without bound as x goes to negative infinity.
Thanks for pointing that out. I'm still stuck on c). Is it simply letting f"(x) = 0?
 
  • #9
HallsofIvy
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"Points of inflection" are places where the derivative changes sign. Those can only occur, of course, at places where the second derivative is 0 but you should then check that the first derivative actually does change sign.
 
  • #10
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"Points of inflection" are places where the derivative changes sign. Those can only occur, of course, at places where the second derivative is 0 but you should then check that the first derivative actually does change sign.
Yeh, for the first derivate at x=0 a minimum occurs and at x=2/a a maximum occurs.

Im having problems solving for f"(x) = 0 though. I take out e^-ax as a common factor but I'm then left with (-2xa) + (x^2a^2) + 2 - 2x = 0

:confused: :confused:
 
  • #11
Dick
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I don't quite agree with your result for f''(x) with the exponential factored out. But once you get it right, then this looks like a job for the quadratic formula, doesn't it?
 
  • #12
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I don't quite agree with your result for f''(x) with the exponential factored out. But once you get it right, then this looks like a job for the quadratic formula, doesn't it?
I just double checked and it still seems right to me. Can you suggest what you think is wrong?
 
  • #13
Dick
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I get a -4*a*x and no 2*x term.
 
  • #14
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I get a -4*a*x and no 2*x term.
yeah, you are right, the an a went missing in the -2*x term which would have made it -4xa by grouping with the other -2xa.

so after all that I get...
http://img264.imageshack.us/img264/3452/untitledxu2.jpg [Broken]
somebody hinted to use quadratic formula, so a= a^2, b= -4x, c =2 etc etc

figure all that out to get...
http://img413.imageshack.us/img413/948/quadformxq7.jpg [Broken]

correct? (crosses fingers)
 
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  • #15
Dick
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You're getting there. Do you mean (sqrt(2))*a or sqrt(2*a). The former, I hope.
 
  • #16
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You're getting there. Do you mean (sqrt(2))*a or sqrt(2*a). The former, I hope.
Yeh, just extended the roof over a in my working...
silly error :yuck:

Now for part d). I'm really puzzled as to how you would prove something like this mathematically especially given the unknown a.
 
  • #17
Dick
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You have tons of information about the function. Sketch a graph and label significant points in terms of a. Also examine the limiting behavior of the function as x->+/- infinity.
 
  • #18
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I can't figure out the limiting behaviour as x-> - infinity.

x^2 will become infinity, but at the same time e^-(ax) will become a very small number. In graphical terms, what does that mean?
 
  • #19
Dick
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What kind of a number is -a*x if a>0 and x<0? I don't think e^(-ax) gets small at all. If you mean x->+infinity do you know L'Hopital's rule (judging from the problem you are working on you might not).
 
  • #20
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What kind of a number is -a*x if a>0 and x<0? I don't think e^(-ax) gets small at all. If you mean x->+infinity do you know L'Hopital's rule (judging from the problem you are working on you might not).
oh, my bad! I don't know that rule, can you please explain?
 
  • #21
Dick
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We want the lim (x->infinity) of e^(-ax)*x^2. Rewrite this as x^2/e^(ax). Now it has the form infinity/infinity. L'Hopital says I can take the derivative of numerator and denominator. Get 2*x/(a*e^(ax)). Still infinity/infinity. Do it again. 2/(a^2*e^(ax)). Now the limit is clearly zero. So that's the original limit.

If you don't know tricks like that you can probably just eyeball it. What would you guess to be the value of something like (100)^2*e(-100). Really big or really small?
 
  • #22
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If you don't know tricks like that you can probably just eyeball it. What would you guess to be the value of something like (100)^2*e(-100). Really big or really small?

Very small. So now I know:
lim -> -infinity = -infinity because x²e^(-ax) ->0 and
lim->infinity = 0 because x²e^(-ax) ->0

But graphically, im not too sure whats going on.

Well I know there is a minimum at x=0 and x=(2/a), and a point of inflexion at x=[2a±√(2)*a]/a² but im not too sure about the asymptotes and global max/min.

I know this is probably very wrong, but no harm trying..

http://img239.imageshack.us/img239/6922/graphfb1.jpg [Broken]
 
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  • #23
Dick
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Your graph is completely correct! So global max, global min?
 
  • #24
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Your graph is completely correct! So global max, global min?
LOL!

Global max at x=-infinity
global min at x=0

Just a question regarding graph protocol.
With that asymptote at y=0, is it correct to just have the dotted line towards the end?

Also, am I missing an verticle asymptote at x=-infinity? :cool:
 
  • #25
Dick
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You can draw it anyway that makes it clear to you. But most people would not say that there is a global max at -infinity since it's not really a point on the line. Similarly, no vertical asymptote there.
 

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