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Very simple derivative problem don't know what I'm doing wrong!

  • #1
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Case closed!

Homework Statement



I'm reviewing for a test and I'm surprised of this. I've did countless problems harder than this.. for some reason every time I use the constant rule I get it wrong.. I want to understand.. WHY?!?

Determine the points at which the graph of the function has a horizontal line.

[tex]g(x)=\frac{8(x-2)}{e^x}[/tex]

The Attempt at a Solution



[tex]g(x)=\frac{8(x-2)}{e^x}[/tex]
By the constant rule:
[tex]=8 (\frac{x-2}{e^x}[/tex]
Definition of quotient rule with constant
[tex]c \frac{\frac{d}{dx} f(x) g(x) - \frac{d}{dx} g(x) f(x)}{[g(x)]^2}[/tex]
[tex] 8 \frac{1(e^x)-e^x(x-2)}{e^{2x}}[/tex]
Distribute [tex]e^x[/tex]
[tex]=8 \frac{e^x - xe^x - 2e^x}{e^{2x}}[/tex]
Divide by [tex]e^x[/tex]
[tex]= 8 \frac{-x-2}{e^x}[/tex]
[tex]= \frac{-8x-16}{e^x}[/tex]
Horizontal line (0 slope) is at g'(-2)
[tex] g(2)= \frac{-8(x-2}{e^x}[/tex]
[tex] g(2) = \frac{-32}{e^-2}[/tex]
[tex] g(2) = -32e^2[/tex]
So g(x) has a horizontal tangent at [tex](-2,-32e^2)[/tex]

But the right answer is supposed to be horizontal line slope at g'(3) and horizontal tangent at [tex](3,\frac{8}{e^3})[/tex]

What am I doing wrong? Am I using the constant rule the wrong way? It makes me go crazy because I know I have all the basic derivative rules in check. Please help!
 
Last edited:

Answers and Replies

  • #2
SammyS
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Homework Statement



Determine the points at which the graph of the function has a horizontal line.

[tex]g(x)=\frac{8(x-2)}{e^x}[/tex]

The Attempt at a Solution



[tex]g(x)=\frac{8(x-2)}{e^x}[/tex]
...

Distribute [tex]e^x[/tex]
[tex]=8 \frac{e^x - xe^x - 2e^x}{e^{2x}}[/tex]
It's OK to here.
Divide by [tex]e^x[/tex]
[tex]= 8 \frac{-x-2}{e^x}[/tex]
,,,
[itex]\displaystyle \frac{e^x}{e^x}-\frac{xe^x}{e^x}-\frac{2e^x}{e^x}=1-x+2\,.[/itex]
 
  • #3
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It's OK to here.


[itex]\displaystyle \frac{e^x}{e^x}-\frac{xe^x}{e^x}-\frac{2e^x}{e^x}=1-x+2\,.[/itex]
Hey, I've edited a portion of my post it wasn't done yet. But I am a bit confused about what you just wrote. Don't you mean:
[itex]\displaystyle \frac{e^x}{e^x}-\frac{xe^x}{e^x}-\frac{2e^x}{e^x}=1-x-2\,.[/itex]

Edit: Either way.. thanks!! I will edit my post and make the correction accordingly.
 
  • #4
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It's OK to here.


[itex]\displaystyle \frac{e^x}{e^x}-\frac{xe^x}{e^x}-\frac{2e^x}{e^x}=1-x+2\,.[/itex]
I don't understand how you got +2. I re-ran my calculation as [tex]=1-x-2[/tex]. But it seems that I only get the right answer if I put [tex]=1-x+2[/tex]

Could you explain how its +2 and not -2? This is really testing my faith in my algebra foundation right now..
 
  • #5
Deveno
Science Advisor
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6
[tex]-e^x(x - 2) = -e^x(x) -e^x(-2) = -xe^x + 2e^x[/tex]
 
  • #6
SammyS
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...
[tex]8 \frac{1(e^x)-e^x(x-2)}{e^{2x}}[/tex]

Distribute [tex]e^x[/tex]
[tex]=8 \frac{e^x - xe^x - 2e^x}{e^{2x}}[/tex]
My earlier post was in error.

Distributing the ex gives a numerator of:

ex - xex + 2ex

Then dividing this by ex gives: 1 - x + 2 = 3 - x .
 
  • #7
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0o silly me.. It ended up being a distributive mistake all along 0__O. Hahaha.. I must need a break then.. I find that quite funny. Well thanks for pointing the mistake. :approve:
[tex]-e^x(x - 2) = -e^x(x) -e^x(-2) = -xe^x + 2e^x[/tex]
My earlier post was in error.

Distributing the ex gives a numerator of:

ex - xex + 2ex

Then dividing this by ex gives: 1 - x + 2 = 3 - x .
 

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