for the following function, $f(x)=e^{-x}+1$, find\na. the domain and range.\ndomain:\nrange:\nb. the…

Explanation:

Step1: Find the domain

The exponential function (y = e^{-x}) is defined for all real values of (x). So, for the function (f(x)=e^{-x}+1), the domain is all real numbers. In interval notation, the domain is ((-\infty,\infty)) since there are no restrictions on the input (x) (no values of (x) make the function undefined).

Step2: Analyze the range

We know that the exponential function (y = e^{-x}=\left(\frac{1}{e}\right)^{x}) has a range of ((0,\infty)) because (e^{-x}>0) for all (x\in R). Then, if (y = e^{-x}+1), we add (1) to each value of (e^{-x}). Using the property of inequalities, if (e^{-x}>0), then (e^{-x}+1>0 + 1). So the range of (y=e^{-x}+1) is ((1,\infty))

Step3: Determine the horizontal asymptote

For an exponential function of the form (y = a\cdot e^{bx}+c), the horizontal asymptote is (y = c). In the function (f(x)=e^{-x}+1), comparing with (y = a\cdot e^{bx}+c) ((a = 1), (b=- 1), (c = 1)), as (x\rightarrow\infty), (e^{-x}=\frac{1}{e^{x}}\rightarrow0). So (y=e^{-x}+1\rightarrow0 + 1). The equation of the horizontal asymptote is (y = 1)

Answer:

Domain: ((-\infty,\infty)) Range: ((1,\infty)) Horizontal asymptote: (y = 1)