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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!

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