The curve with equation $y = \frac {e^{2x}} {4 + e^{3x}}$ has one stationary point. Find the exact values of the coordinates of this point.
I got to the point where this is my $\frac {dy} {dx}$:$$\frac{ (4 + e^{3x}) (2e{^{2x}})-e^{2x}(3e^{3x})}{(4+e^{3x})^2} = 0$$
Is this correct to find the stationary points, if it is, how do I get $x $ from this equation?
Best Answer
Developing your numerator, that must be zero, you have the following: $$ 2e^{5x} + 8e^{2x} - 3e^{5x} = 0 \iff e^{5x} = 8e^{2x} \iff 8 = \frac{e^{5x}}{e^{2x}} = e^{3x} \iff \ln 8 = \ln e^{3x} = 3x\ln e = 3x \iff x = \frac{\ln 8}{3} = \frac{\ln2^3}{3} = \frac{3\ln 2}{3} = \ln 2 $$ Notice that you can divide by an exponential because $e^x\neq 0\;\;\forall x\in\mathbb{R}$.