# Critical numbers

• Apr 8th 2008, 11:10 AM
XIII13Thirteen
Critical numbers
Need a bit of help

Determine

a.) All the critical numbers

b.) all relative extrema

$y(x)=(x-1)e^{2x}$

I'm thinking my derivative is ...

$2*(x-1)*(1)*e^{2x}$

Multiple choice:

a.){ $\frac{1}{2}$,max}

b.){0,min}

c.) { $\frac{1}{2}$, min}

d.) none of these

I'm thinking it is $\frac{1}{2}$ min
• Apr 8th 2008, 11:13 AM
Jhevon
Quote:

Originally Posted by XIII13Thirteen
Need a bit of help

Determine

a.) All the critical numbers

b.) all relative extrema

$y(x)=(x-1)e^{2x}$

I'm thinking my derivative is ...

$2*(x-1)*(1)*e^{2x}$

Multiple choice:

a.){ $\frac{1}{2}$,max}

b.){0,min}

c.) { $\frac{1}{2}$, min}

d.) none of these

I'm thinking it is $\frac{1}{2}$ min

you derivative is wrong. you need the product rule here.
• Apr 8th 2008, 11:16 AM
XIII13Thirteen
$(x-1)2e^{2x}+(1)e^{2x}$?
• Apr 8th 2008, 11:20 AM
Moo
Hello,

Yes, it's it ;)

$(x-1)2e^{2x}+(1)e^{2x}=e^{2x}(2x-2+1)=e^{2x}(2x-1)$
• Apr 8th 2008, 11:22 AM
Jhevon
Quote:

Originally Posted by XIII13Thirteen
$(x-1)2e^{2x}+(1)e^{2x}$?

very good. and Moo was kind enough to factorize it for us. now set Moo's factorized form to zero and solve for your critical points.
• Apr 8th 2008, 11:24 AM
XIII13Thirteen
Quote:

Originally Posted by Jhevon
very good. and Moo was kind enough to factorize it for us. now set Moo's factorized form to zero and solve for your critical points.

Yes, x=1/2 will zero out the equation. I do believe X>0 is a minimum? If so than c. Thanks guys