Question

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:

\(0.2254\)