1. ## 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

2. 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.

3. $(x-1)2e^{2x}+(1)e^{2x}$?

4. Hello,

Yes, it's it

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

5. 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.

6. 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