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Finding slope locations

  • Thread starter Johnyi
  • Start date
  • #1
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Homework Statement



Let f(x) = xe^(2x)

A) Find the values of x for which the slope of the cruve y = f(x) is 0
B) Explain the meaning of your answer to part (a) in terms of the graph f


Homework Equations





The Attempt at a Solution



I tried setting xe^(2x) to 0 by making it 0 = xe^(2x). I dont know what to do from there on. Am i supposed to take the derivative of xe^(2x)?
 

Answers and Replies

  • #2
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A derivative is a fancy way of saying "instantaneous slope." So if you need to find at which values of x the slope is instantaneously 0, you need to set the derivative to zero, not the function itself.
 
  • #3
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A derivative is a fancy way of saying "instantaneous slope." So if you need to find at which values of x the slope is instantaneously 0, you need to set the derivative to zero, not the function itself.

So i get 2xe^(2x) = 0
The answers in the back of the book say that x = 1/2 but i just cant seem to find a way to that answer
 
  • #4
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So i get 2xe^(2x) = 0
The answers in the back of the book say that x = 1/2 but i just cant seem to find a way to that answer
Product rule.
 
  • #5
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Product rule.
Im sorry i just cant seem to do this right..so far i got

x(2e^(2x))+e^(2x)
=2e^(2x)x+e^(2x)
 
  • #6
280
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Im sorry i just cant seem to do this right..so far i got

x(2e^(2x))+e^(2x)
=2e^(2x)x+e^(2x)
Yes, that looks right. Now set that equal to zero and solve for x. For the record, as you've written f(x), I believe the answer in the book is incorrect.
[tex] 2xe^{2x}+e^{2x} = 0[/tex]

Can you solve it from here? What answer do you find? I will tell you if I find the same answer.
 
  • #7
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I thinking that you can factor out a e^(2x).

So it would be e^(2x)(x+1).

Then subtract 1 to the other side, then divide by...I have no clue
 
  • #8
280
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I thinking that you can factor out a e^(2x).

So it would be e^(2x)(x+1).

Then subtract 1 to the other side, then divide by...I have no clue
if A(x)B(x) = 0, what must be true about at least A(x) or B(x)? Think back to algebra I where you solved

[tex]x^2-1=(x+1)(x-1)=A(x)B(x) = 0[/tex]
[tex]A(x)=x+1[/tex]
[tex]B(x)=x-1[/tex]

Another method to arriving to the same conclusion is:
What does
[tex] \frac{0}{A}[/tex]
equal if A is nonzero? Is an exponential ever zero? A/A = 1 if A is nonzero too. Can you think of a single step to remove the exponential?
 
Last edited:
  • #9
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Im sorry i dont understand
 
  • #10
280
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Im sorry i dont understand
Divide both sides by the exponential or you can simply say if A(x)B(x) = 0, A(X) = 0 or B(x) = 0.
[tex] A(x) = e^{2x}[/tex]
[tex] B(x) = 2x+1[/tex]
A(x) cannot reach zero for a finite x. B(x) can. So solve for when B(x) = 0 for the solution. Note, you end up solving for B(x) = 0 if you divide both sides by A(x) (the exponential).
 
Last edited:

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