1. The problem is to analyze the function $$y = e^{2\sqrt{2x}}$$. 2. This is an exponential function where the exponent is $$2\sqrt{2x}$$. The base $$e$$ is Euler's number, approximately 2.718. 3. Important rules: - The square root function $$\sqrt{2x}$$ is defined only for $$x \geq 0$$. - The exponential function $$e^u$$ is always positive for any real number $$u$$. 4. Let's examine the domain: - Since $$\sqrt{2x}$$ requires $$2x \geq 0$$, we have $$x \geq 0$$. 5. Behavior of the function: - When $$x=0$$, $$y = e^{2\sqrt{0}} = e^0 = 1$$. - As $$x$$ increases, $$\sqrt{2x}$$ increases, so the exponent $$2\sqrt{2x}$$ increases. - Therefore, $$y$$ grows exponentially as $$x$$ increases. 6. Summary: - Domain: $$x \geq 0$$. - Range: $$y > 0$$. - The function starts at $$y=1$$ when $$x=0$$ and increases without bound as $$x$$ increases. Final answer: The function $$y = e^{2\sqrt{2x}}$$ is defined for $$x \geq 0$$ and increases exponentially from 1 upwards.