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

解答

切线

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

解答

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}

求解步骤

计算在一般点

x = a_{0}

的切线

找到切点:

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

计算

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

的斜率:

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{6}{, ( 0 , 3 )} (- \frac{1}{( 1 + e ^{- a_{0}} ) ^{2}} (- e^{- a_{0}}))

找到斜率 m=

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

且穿过

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

的直线:

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}

作图

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