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

The tangent of f(x)=(1+x)/(1+e^x),\at x=0 is y= 1/4 x+1/2

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

tangent of

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

y = \frac{1}{4} x + \frac{1}{2}

Solution steps

Find the tangent point:

(0, \frac{1}{2})

Find the slope of

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

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{1}{4}

Find the line with slope m=

\frac{1}{4}

and passing through

(0, \frac{1}{2}) : y = \frac{1}{4} x + \frac{1}{2}

y = \frac{1}{4} x + \frac{1}{2}

\frac{10}{s ( s + 3 )}

2 x^{2} - 6 x

x \to 2 + lim (∣ x + 2 ∣ - 2)

\frac{s}{s - 3}

\frac{d}{d x} (4 x - 3 ln (x))

What is the tangent of f(x)=(1+x)/(1+e^x),\at x=0 ?
The tangent of f(x)=(1+x)/(1+e^x),\at x=0 is y= 1/4 x+1/2