What is the tangent of f(x)=(10)/(1+e^{-x)},\at x=0 ?

The tangent of f(x)=(10)/(1+e^{-x)},\at x=0 is y= 5/2 x+5

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tangent of f(x)=(10)/(1+e^{-x)},\at x=0

tangent of

f (x) = \frac{10}{1 + e ^{- x}}, at x = 0

y = \frac{5}{2} x + 5

Solution steps

Find the tangent point:

(0, 5)

Find the slope of

f (x) = \frac{10}{1 + e ^{- x}} : \frac{df ( x )}{d x} = \frac{10 e ^{- x}}{( 1 + e ^{- x} ) ^{2}}

Compute the slope of

f (x) = \frac{10}{1 + e ^{- x}}

x = 0 : m = \frac{5}{2}

Find the line with slope m=

\frac{5}{2}

and passing through

(0, 5) : y = \frac{5}{2} x + 5

y = \frac{5}{2} x + 5

n = 2 \sum \infty (\frac{n + 1}{n})^{n}

\int \frac{5 ^{x}}{1 + 5 ^{x}} d x

\frac{d}{d x} ((1000 + x)^{- \frac{1}{3}})

\int csc^{2} (x) cot^{2} (x) d x

\int 6 x^{2} tan (x^{3}) d x

What is the tangent of f(x)=(10)/(1+e^{-x)},\at x=0 ?
The tangent of f(x)=(10)/(1+e^{-x)},\at x=0 is y= 5/2 x+5