let $f(x)=\\frac{0.5}{1 + 9e^{-2x}}$. evaluate $f(1)$ to 4 decimal places. $f(1)=$

Explanation:

Step1: Substitute (x = 1) into the function

Substitute (x = 1) into (f(x)=\frac{0.5}{1 + 9e^{-2x}}), we get (f(1)=\frac{0.5}{1+9e^{-2\times1}}=\frac{0.5}{1 + 9e^{-2}}).

Step2: Calculate the value of (e^{-2})

We know that (e\approx2.71828), so (e^{-2}=\frac{1}{e^{2}}\approx\frac{1}{(2.71828)^{2}}\approx\frac{1}{7.38906}\approx0.13534).

Step3: Calculate the denominator

The denominator (1 + 9e^{-2}\approx1+9\times0.13534=1 + 1.21806=2.21806).

Step4: Calculate the value of (f(1))

(f(1)=\frac{0.5}{2.21806}\approx0.2254).

Answer:

(0.2254)