1. **State the problem:** We are given the function $$y=1-2e^{\frac{1}{2}x+1}$$ and asked to identify the basic function, sketch it over the interval $$0 \leq x \leq 1$$, identify shifting and scaling effects, apply these effects, and write the domain and range. 2. **Identify the basic function:** The basic function here is the exponential function $$f(x) = e^x$$. 3. **Rewrite the given function:** The exponent can be rewritten as $$\frac{1}{2}x + 1 = \frac{1}{2}x + 1$$, so the function is $$y = 1 - 2e^{\frac{1}{2}x + 1}$$. 4. **Identify transformations:** - The term $$e^{\frac{1}{2}x}$$ indicates a horizontal scaling by a factor of 2 (since $$x$$ is multiplied by $$\frac{1}{2}$$). - The $$+1$$ in the exponent corresponds to a horizontal shift to the left by 2 units because $$e^{x+c} = e^c e^x$$, so $$e^{\frac{1}{2}x + 1} = e^1 e^{\frac{1}{2}x}$$ which is a vertical scaling by $$e^1 = e$$. - The multiplication by $$-2$$ outside the exponential reflects the graph vertically and scales it by 2. - The addition of $$+1$$ outside shifts the graph vertically upward by 1. 5. **Express the function with transformations:** $$y = 1 - 2e^{1} e^{\frac{1}{2}x} = 1 - 2e \cdot e^{\frac{1}{2}x}$$ 6. **Domain:** The domain of an exponential function is all real numbers, so $$\text{Domain} = (-\infty, \infty)$$. 7. **Range:** Since $$e^{\frac{1}{2}x} > 0$$ for all $$x$$, the term $$-2e e^{\frac{1}{2}x}$$ is always negative and decreases without bound as $$x$$ increases. - The maximum value of $$y$$ occurs when $$x \to -\infty$$, where $$e^{\frac{1}{2}x} \to 0$$, so $$y \to 1$$ from below. - The function decreases without bound as $$x \to \infty$$. Thus, $$\text{Range} = (-\infty, 1)$$. 8. **Sketch over $$0 \leq x \leq 1$$:** - At $$x=0$$: $$y = 1 - 2e e^{0} = 1 - 2e \approx 1 - 2 \times 2.718 = 1 - 5.436 = -4.436$$ - At $$x=1$$: $$y = 1 - 2e e^{\frac{1}{2}} = 1 - 2e e^{0.5} = 1 - 2e^{1.5} \approx 1 - 2 \times 4.4817 = 1 - 8.9634 = -7.9634$$ The graph decreases steeply from about $$-4.436$$ at $$x=0$$ to about $$-7.9634$$ at $$x=1$$.