Find the maximum and minimum values of:

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Homework Help Overview

The problem involves finding the maximum and minimum values of the function f(x) = x^4e^-x over the interval [0,10]. Participants are discussing the process of determining critical points and evaluating the function at specific values.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss setting the derivative f' equal to zero to find critical points, with some attempting to factor the derivative. There are questions about the implications of setting e^-x to zero and the interpretation of logarithmic expressions.

Discussion Status

The discussion is active, with participants exploring different interpretations of the critical points and the behavior of the function. Some guidance has been offered regarding the factoring of the derivative and the nature of the roots, but no consensus has been reached on the complete analysis of the function.

Contextual Notes

Participants are working within the constraints of the specified interval [0,10] and are questioning the validity of certain mathematical expressions related to the roots of the derivative.

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Homework Statement



Find the maximum and minimum values of

f(x)=x^4e^-x over [0,10]


Homework Equations





The Attempt at a Solution



From what I remember is you set the f' equal to 0. And that will give you two more points to test. and then you put the 0 and 10 number into the equation to find the values.

When I put 0 or 10 in i get 0 for both.

f'= (4x^3)(e^-x)+(x^4)(-e^-x)

If i set that equal to 0; I really do not know how to find the what x is.
 
Last edited:
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Well, first, let's write this out...

[tex]f'(x) = 4 x^3 e^{-x} - x^4 e^{-x} = 0[/tex]

First, I'd recommend factoring.
 
okay i see that now. so you get something like:

(x3)(e-x)[4-x]

so then

(x3)(e-x)=0

and (4-x)=0

so you get x=4

and then i think the other one is ln0/-1 which is undefined if you set it to 0.
 
Yep, okay i got it. Maximum is 4.6888 at x=4. and Min is 0 at x=0

I don't know if (x3)(e^-x)=0 gives you ln0/-1 but either way 4 is the value I was looking for.

Thank You
 
Last edited:
Hmm... I'm not sure where you're getting -log(0)... You know that if [itex]x^3 e^{-x} = 0[/itex], then either x^3=0 or e^(-x)=0, and only one of those has a solution.

You've found a critical point at x=4, but there's one more.
 
If you set x^3=0 You get x=0?
If you set e^-x=0 You multiply both by (ln) so -x =ln(0) divide both by -1. That is what I thought? ln(0) / -1
 
Yes, but since [itex]ln(0)=-\infty[/tex], what you essentially said is that e^(-x)=0 for an x outside the standard real (and complex, incidentally) number system. So, it doesn't count as a root.<br /> <br /> Also, your domain was [0,10]. Negative infinity is outside that domain.[/itex]
 
I would be inclined to write [itex]fsingle-quote(x) = 4 x^3 e^{-x} - x^4 e^{-x} = 0[/itex] as [itex]4x^3e^{-x}= x^4e^{-x}[/itex]. Since [itex]e^{-x}[/itex] is never 0, we can divide both sides of the equation by it and get [math]4x^3= x^4[/math]. If x is not 0, we can divide both sides by x to get x= 4.

That tells you the two roots of the equation.
 

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