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

The tangent of f(x)= 1/(x+1),\at x=0 is y=-x+1

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

tangent of

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

y = - x + 1

Solution steps

Find the tangent point:

(0, 1)

Find the slope of

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

Compute the slope of

f (x) = \frac{1}{x + 1}

x = 0 : m = - 1

Find the line with slope m=

- 1

and passing through

(0, 1) : y = - x + 1

y = - x + 1

\int 2 cos (x) sin^{2} (x) d x

x + 1

\frac{4 s + 7}{s ^{2} - 4}

\int \frac{x ^{2}}{x ^{2} + 1} d x

(y + 1) (\frac{d y}{d x}) = 2 x

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