I said

f'(x) = m = 0 = 2xe^2x ---> x = 0/(2xe^2x) = 0

Is this right?

- Apr 4th 2010, 08:30 AMsatxFor what x value does the tangent of the graph of f(x) = xe^2x have a slope = 0?
I said

f'(x) = m = 0 = 2xe^2x ---> x = 0/(2xe^2x) = 0

Is this right? - Apr 4th 2010, 08:37 AMe^(i*pi)
Your method is right although you need to use the product rule when finding $\displaystyle f'(x)$

$\displaystyle f(x) = xe^{2x}$

$\displaystyle u = x \: \rightarrow \: u' = 1$

$\displaystyle v = e^{2x} \: \rightarrow \: v' = 2e^{2x}$

$\displaystyle f'(x) = u'v + v'u$ - Apr 4th 2010, 08:41 AMsatx
Wouldn't that equal 2?

- Apr 4th 2010, 08:50 AMRaoh
- Apr 4th 2010, 08:53 AMRaoh
to solve $\displaystyle f'(x)=0$

$\displaystyle f'(x)=2e^{2x}x+e^{2x}=e^{2x}(2x+1)=0\Leftrightarro w 2x+1=0$

:) - Apr 4th 2010, 08:54 AMsatx
Ok then that would equal one then, if x = 0, no? (e^2(0))(0) + e^2(0)) = 0 + 1 = 1???

- Apr 4th 2010, 08:55 AMsatx
- Apr 4th 2010, 08:56 AMRaoh
- Apr 4th 2010, 08:58 AMsatx
- Apr 4th 2010, 09:05 AMRaoh
- Apr 4th 2010, 09:15 AMRaoh