Express function in terms of hyperbolic function.

[JasonFMH]

Express function f(x) given in terms of a hyperbolic function.

f(x) = e^x / (e^(2x)+1)

The first part of the problem was finding the average value. However, I don't know how to change a function to a hyperbolic function. Any sort of hints toward getting me started would be greatly appreciated.

 

[galactus]

\(\displaystyle \frac{e^{x}}{e^{2x}+1}=\frac{1}{2}sech(x)\)

Note that \(\displaystyle cosh(x)=\frac{1}{2}(e^{x}+e^{-x})\)

 

[JasonFMH]

I'm sorry about asking another question, but we didn't really touch bases with this in class. Could you give me a very brief run down on how exactly to show it. Is it just memorization, much like the inverse trig functions and then seeing which applies to that particular function?

 

[Deleted member 4993]

I'm sorry about asking another question, but we didn't really touch bases with this in class. Could you give me a very brief run down on how exactly to show it. Is it just memorization, much like the inverse trig functions and then seeing which applies to that particular function?

You should derive it from the definition that Galactus gave you:

\(\displaystyle cosh(x)\ \ = \frac{e^x + e^{-x}}{2}\)

 

[BigGlenntheHeavy]

\(\displaystyle f(x) \ = \ \frac{e^x}{e^{2x}+1} \ = \ \frac{1}{e^{-x}(e^{2x}+1)} \ = \ \frac{1}{e^x+e^{-x}}\)

\(\displaystyle Now, \ cosh(x) \ = \ \frac{e^x+e^{-x}}{2}, \ \implies \ sech(x) \ = \ \frac{2}{e^x+e^{-x}}\)

\(\displaystyle Hence, \ 2f(x) \ = \ \frac{2}{e^x+e^{-x}} \ = \ sech(x).\)

\(\displaystyle Therefore, \ f(x) \ = \ \frac{1}{e^x+e^{-x}} \ = \ \frac{1}{2}sech(x).\)