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

1. ## For 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?

2. 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 $f'(x)$

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

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

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

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

3. Wouldn't that equal 2?

4. Originally Posted by satx
Wouldn't that equal 2?
i believe you have, $f'(x)=2e^{2x}x+e^{2x}$

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

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

7. 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

8. 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 $f'(x)=0$

9. Originally Posted by Raoh
i think you're trying to solve $f'(x)=0$
Well isn't the slope equal to the derivative?

10. 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 $-\frac{1}{2}$.

11. Originally Posted by satx
Well isn't the slope equal to the derivative?
yes