What is the tangent of f(x)= 6/((1+e^{-x)),\at (0,3)} ?

The tangent of f(x)= 6/((1+e^{-x)),\at (0,3)} is

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tangent of f(x)= 6/((1+e^{-x)),\at (0,3)}

tangent of

f (x) = \frac{6}{( 1 + e ^{- x} ) , at ( 0 , 3 )}

y = \frac{6}{, ( 0 , 3 )} (- \frac{1}{( 1 + e ^{- a_{0}} ) ^{2}} (- e^{- a_{0}})) x + \frac{6}{( 1 + e ^{- a_{0}} ) , ( 0 , 3 )} - \frac{6}{, ( 0 , 3 )} (- \frac{1}{( 1 + e ^{- a_{0}} ) ^{2}} (- e^{- a_{0}})) a_{0}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{6}{( 1 + e ^{- a_{0}} ) , ( 0 , 3 )})

Find the slope of

f (x) = \frac{6}{( 1 + e ^{- x} ) , ( 0 , 3 )} : \frac{df ( x )}{d x} = \frac{6}{, ( 0 , 3 )} (- \frac{1}{( 1 + e ^{- x} ) ^{2}} (- e^{- x}))

Compute the slope of

f (x) = \frac{6}{( 1 + e ^{- x} ) , ( 0 , 3 )}

x = a_{0} : m = \frac{6}{, ( 0 , 3 )} (- \frac{1}{( 1 + e ^{- a_{0}} ) ^{2}} (- e^{- a_{0}}))

Find the line with slope m=

\frac{6}{, ( 0 , 3 )} (- \frac{1}{( 1 + e ^{- a_{0}} ) ^{2}} (- e^{- a_{0}}))

and passing through

(a_{0}, \frac{6}{( 1 + e ^{- a_{0}} ) , ( 0 , 3 )}) : y = \frac{6}{, ( 0 , 3 )} (- \frac{1}{( 1 + e ^{- a_{0}} ) ^{2}} (- e^{- a_{0}})) x + \frac{6}{( 1 + e ^{- a_{0}} ) , ( 0 , 3 )} - \frac{6}{, ( 0 , 3 )} (- \frac{1}{( 1 + e ^{- a_{0}} ) ^{2}} (- e^{- a_{0}})) a_{0}

y = \frac{6}{, ( 0 , 3 )} (- \frac{1}{( 1 + e ^{- a_{0}} ) ^{2}} (- e^{- a_{0}})) x + \frac{6}{( 1 + e ^{- a_{0}} ) , ( 0 , 3 )} - \frac{6}{, ( 0 , 3 )} (- \frac{1}{( 1 + e ^{- a_{0}} ) ^{2}} (- e^{- a_{0}})) a_{0}

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What is the tangent of f(x)= 6/((1+e^{-x)),\at (0,3)} ?
The tangent of f(x)= 6/((1+e^{-x)),\at (0,3)} is