Help me with this.
Can you tell me what is your value of x.Thanks
i dunno where went wrong..
Fortunately I have a degree in Egyptology majoring in heiroglyphics. The curve is $\displaystyle y = 3 e^{-x/2} + e^{x/2}$.
Solve $\displaystyle \frac{dy}{dx} = 0$ for x and substitute the value into y:
$\displaystyle 0 = -\frac{3}{2} e^{-x/2} + \frac{1}{2} e^{x/2}$
$\displaystyle \Rightarrow 0 = -3 + e^x \Rightarrow x = \ln 3$.
Therefore $\displaystyle y = 3 e^{-(\ln 3) /2} + e^{(\ln 3)/2} = 3 = 3 e^{\ln \left(1/\sqrt{3}\right)} + e^{\ln \sqrt{3}} = \frac{3}{\sqrt{3}} + \sqrt{3} = 2 \sqrt{3}$.
$\displaystyle 0 = -\frac{3}{2} e^{-x/2} + \frac{1}{2} e^{x/2}$
can you tell me how to get the value of x from here.
Its because i tried to use the substitution method to get x
For eg. Let x=e^x/2
But i got a different answer.Can you show me another method?
P.S.My dy/dx is correct