Exponentials Q. (1 Viewer)

[NewiJapper]

Find the stationary points and the nature of the curve (x+1)e^x.


I can do other questions of this type but for some reason this stumps me :S lol

 

[nutcracker]

Hmm... I'll give it a go, but I'm really bad at maths, so beware of receiving a hideously wrong answer ><;.

okaay.

y = (x+1)e^x

let u = (x+1)
u' = 1
let v = e^x
v' = e^x

y' = vu' + uv'
= e^x + e^x(x+1)
= e^x(x +2)

For stat points let y' = 0

e^x(x+2) = 0

e^x = 0
no solution?

(x+2) = 0
x = -2

when x = -2, y = -e^(-2)


y'' = e^x(x+3)
sub in x = -2
y'' = e^(-2) x (1)
> 0
therefore min point at (-2, -e^-2)

I have a feeling I've done something terribly wrong.
Do you know the answer?

 

[marmsie]

You haven't done anything wrong, that is the correct answer and method.

And yes, e^x = 0 has no solutions.

 

[NewiJapper]

Thanks 1 questions though.

Does it matter if i label u as e^x and v as (x+1)? Because thats how we did it in class.

 

[marmsie]

Not for the product rule as it is adding and multiplying everything so it will not matter the order that you do things in.

But just remember it does matter for the quotient rule because you have to subtract and divide your u's and v's, so if you mix them up you will get a completely different answer.

 

[NewiJapper]

when you "label" them for the quotient rule when its a/b a=u and b=v?

 

[marmsie]

That's right:

<img src="http://www.codecogs.com/eq.latex?\begin{align*}y&=\frac{u}{v}\\y'&= \frac{u'v-v'u}{v^2}\end{align}">