What is f(2) given $f (x) = \frac{6}{1 + 5 e^{- 3 x}}$ $f(x)=6/(1+5e^(-3x))$ ?

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Answer 1 · 2017-10-18T16:52:26

$f (2) = \frac{6}{1 + 5 e^{- 6}} \approx 5.9265477839$ $f(2) = 6/(1+5e^-6) ~~5.9265477839$

Given: $f (x) = \frac{6}{1 + 5 e^{- 3 x}}$ $f(x)=6/(1+5e^(-3x))$

To find f(2), one simply substitutes 2 for x everywhere in the function:

$f (2) = \frac{6}{1 + 5 e^{- 3 (2)}}$ $f(2)=6/(1+5e^(-3(2)))$

A representation of the exact value is:

$f (2) = \frac{6}{1 + 5 e^{- 6}}$ $f(2)=6/(1+5e^(-6))$

One can use a calculator to obtain an approximate value:

$f (2) \approx 5.9265477839$ $f(2) ~~5.9265477839$

What is f(2) given f(x)=61+5e−3xf(x)=6/(1+5e^(-3x))?