Answer
a) $y=\frac{1}{2}x+1$
b) See graph
Work Step by Step
a) To get the equation of the line tangent to the graph, we need to know the slope, which we can find by taking the derivative. I also rearranged the equation here to avoid the quotent rule in favor of the chain rule, which is often easier to compute.$$f(x)=\frac{2}{1+e^{-x}}=2(1+e^{-x})^{-1}$$
$$f'(x)=2\frac{d}{dx}(1+e^{-x})^{-1}$$
Applying the chain rule twice, we get:
$$f'(x)=2\times-1\times(1+e^{-x})^{-2}\times-e^{-x}$$
$$=\frac{2}{e^x(1+e^{-x})^2}$$
Now we need to plug in $x=0$ to find the slope at that point.
$$\frac{2}{1(1+1)^2}=\frac{2}{4}=\frac{1}{2}$$
Using the coordinate $(0,1)$ and the slope we found, we can determine that the line tangent to the graph at that point is $y-1=\frac{1}{2}(x-0)$ (or $y=\frac{1}{2}x+1$).
b) See graph (tangent line is in red and the graph is in blue).
