Critical numbers

**XIII13Thirteen** · April 8th 2008, 11:10 AM

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

**Jhevon** · April 8th 2008, 11:13 AM

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.

**XIII13Thirteen** · April 8th 2008, 11:16 AM

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

**Moo** · April 8th 2008, 11:20 AM

Hello,

Yes, it's it

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

**Jhevon** · April 8th 2008, 11:22 AM

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.

**XIII13Thirteen** · April 8th 2008, 11:24 AM

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

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