For what x value does the tangent of the graph of f(x) = xe^2x have a slope = 0?

**satx** · April 4th 2010, 08:30 AM

I said

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

Is this right?

**e^(i*pi)** · April 4th 2010, 08:37 AM

Originally Posted by satx

I said

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

Is this right?

Your method is right although you need to use the product rule when finding $f'(x)$

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

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

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

$f'(x) = u'v + v'u$

**satx** · April 4th 2010, 08:41 AM

Wouldn't that equal 2?

**Raoh** · April 4th 2010, 08:50 AM

Originally Posted by satx

Wouldn't that equal 2?

i believe you have, $f'(x)=2e^{2x}x+e^{2x}$

**Raoh** · April 4th 2010, 08:53 AM

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

**satx** · April 4th 2010, 08:54 AM

Ok then that would equal one then, if x = 0, no? (e^2(0))(0) + e^2(0)) = 0 + 1 = 1???

**satx** · April 4th 2010, 08:55 AM

Originally Posted by Raoh

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

So e^0 is 0? I'm so confused

**Raoh** · April 4th 2010, 08:56 AM

Originally Posted by satx

Ok then that would equal one then, if x = 0, no? (e^2(0))(0) + e^2(0)) = 0 + 1 = 1???

i think you're trying to solve $f'(x)=0$

**satx** · April 4th 2010, 08:58 AM

Originally Posted by Raoh

i think you're trying to solve $f'(x)=0$

Well isn't the slope equal to the derivative?

**Raoh** · April 4th 2010, 09:05 AM

Originally Posted by satx

So e^0 is 0? I'm so confused

well $e^0=1$ ,what i wrote means $f'(-\frac{1}{2})=0$ and the $x$ you're searching for is $-\frac{1}{2}$ .

**Raoh** · April 4th 2010, 09:15 AM

Originally Posted by satx

Well isn't the slope equal to the derivative?

yes

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