I said f'(x) = m = 0 = 2xe^2x ---> x = 0/(2xe^2x) = 0 Is this right?
Follow Math Help Forum on Facebook and Google+
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 $\displaystyle f'(x)$ $\displaystyle f(x) = xe^{2x}$ $\displaystyle u = x \: \rightarrow \: u' = 1$ $\displaystyle v = e^{2x} \: \rightarrow \: v' = 2e^{2x}$ $\displaystyle f'(x) = u'v + v'u$
Wouldn't that equal 2?
Originally Posted by satx Wouldn't that equal 2? i believe you have,$\displaystyle f'(x)=2e^{2x}x+e^{2x}$
to solve $\displaystyle f'(x)=0$ $\displaystyle f'(x)=2e^{2x}x+e^{2x}=e^{2x}(2x+1)=0\Leftrightarro w 2x+1=0$
Ok then that would equal one then, if x = 0, no? (e^2(0))(0) + e^2(0)) = 0 + 1 = 1???
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
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 $\displaystyle f'(x)=0$
Originally Posted by Raoh i think you're trying to solve $\displaystyle f'(x)=0$ Well isn't the slope equal to the derivative?
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 $\displaystyle -\frac{1}{2}$.
Originally Posted by satx Well isn't the slope equal to the derivative? yes
View Tag Cloud