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

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.

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

## 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!

Last edited:

SammyS
Staff Emeritus
Homework Helper
Gold Member

## 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 $$e^x$$
$$= 8 \frac{-x-2}{e^x}$$
,,,

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

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.

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..

Deveno
$$-e^x(x - 2) = -e^x(x) -e^x(-2) = -xe^x + 2e^x$$

SammyS
Staff Emeritus
Homework Helper
Gold Member
...
$$8 \frac{1(e^x)-e^x(x-2)}{e^{2x}}$$

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. $$-e^x(x - 2) = -e^x(x) -e^x(-2) = -xe^x + 2e^x$$

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 .