Finding the stationary point on the curve: $e^x+ye^{-x}=2e^2$

Question

So I'm trying to find the stationary point of this equation: $e^x+ye^{-x}=2e^2$. This is my attempt at doing so:

$$\dfrac{d}{dx}(e^x+ye^{-x})=\dfrac{d}{dx}(2e^2)$$ $$e^x+(-ye^{-x}+\frac{dy}{dx}\times e^{-x})=0$$ $$\dfrac{dy}{dx}=\dfrac{-e^x+ye^{-x}}{e^{-x}}$$ Setting $\dfrac{dy}{dx}=0$ $$-e^x+ye^{-x}=0$$ $$y=\dfrac{e^x}{e^{-x}}=e^{2x}$$

I can't seem to draw a conclusion, and I don't know how to proceed, could anyone help?

Thanks in advance.

Answer 1

Now you have two relations between abscissa and ordinate of stationary point .

$1.)$ The curve itself.

$2.)$ $y=e^{2x}$

Substitute $y=e^{2x}$ in the equation of the curve, you will get $x$-coordinate of stationary point and you know how to calc. $y$-coordinate then.

Answer 2

First find $y$: $$y = 2e^{2+x}-e^{2x}$$ So that: $$y' = 2e^{2+x}-2e^{2x}=0$$ $$e^2=e^x \to x=2$$