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 $\displaystyle f'(x)$

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

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

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

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

3. Wouldn't that equal 2?

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

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

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

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

10. Originally Posted by satx
So e^0 is 0? I'm so confused
well $\displaystyle e^0=1$,what i wrote means $\displaystyle f'(-\frac{1}{2})=0$ and the $\displaystyle x$ you're searching for is $\displaystyle -\frac{1}{2}$.

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