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

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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.

g(x)=\frac{8(x-2)}{e^x}

The Attempt at a Solution



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

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

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!
 
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Nano-Passion said:

Homework Statement



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

g(x)=\frac{8(x-2)}{e^x}

The Attempt at a Solution



g(x)=\frac{8(x-2)}{e^x}
...

Distribute e^x
=8 \frac{e^x - xe^x - 2e^x}{e^{2x}}
It's OK to here.
Divide by e^x
= 8 \frac{-x-2}{e^x}
,,,

\displaystyle \frac{e^x}{e^x}-\frac{xe^x}{e^x}-\frac{2e^x}{e^x}=1-x+2\,.
 
SammyS said:
It's OK to here.\displaystyle \frac{e^x}{e^x}-\frac{xe^x}{e^x}-\frac{2e^x}{e^x}=1-x+2\,.

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:
\displaystyle \frac{e^x}{e^x}-\frac{xe^x}{e^x}-\frac{2e^x}{e^x}=1-x-2\,.

Edit: Either way.. thanks! I will edit my post and make the correction accordingly.
 
SammyS said:
It's OK to here.\displaystyle \frac{e^x}{e^x}-\frac{xe^x}{e^x}-\frac{2e^x}{e^x}=1-x+2\,.
I don't understand how you got +2. I re-ran my calculation as =1-x-2. But it seems that I only get the right answer if I put =1-x+2

Could you explain how its +2 and not -2? This is really testing my faith in my algebra foundation right now..
 
-e^x(x - 2) = -e^x(x) -e^x(-2) = -xe^x + 2e^x
 
Nano-Passion said:
...
8 \frac{1(e^x)-e^x(x-2)}{e^{2x}}

Distribute e^x
=8 \frac{e^x - xe^x - 2e^x}{e^{2x}}

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 .
 
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:
Deveno said:
-e^x(x - 2) = -e^x(x) -e^x(-2) = -xe^x + 2e^x

SammyS said:
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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