What is the tangent of f(x)= 1/(2+x),\at x=2 ?

The tangent of f(x)= 1/(2+x),\at x=2 is y=-1/16 x+3/8

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tangent of f(x)= 1/(2+x),\at x=2

tangent of

f (x) = \frac{1}{2 + x}, at x = 2

y = - \frac{1}{16} x + \frac{3}{8}

Solution steps

Find the tangent point:

(2, \frac{1}{4})

Find the slope of

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

Compute the slope of

f (x) = \frac{1}{2 + x}

x = 2 : m = - \frac{1}{16}

Find the line with slope m=

- \frac{1}{16}

and passing through

(2, \frac{1}{4}) : y = - \frac{1}{16} x + \frac{3}{8}

y = - \frac{1}{16} x + \frac{3}{8}

\int (\frac{12}{x} + 12 x) d x

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

\int (8 x + 3 sin (x))^{2} d x

\int (\frac{1}{4 - 25 x ^{2}}) d x

\frac{14 x + 4 x ^{2}}{( 7 + 4 x ) ^{2}}

What is the tangent of f(x)= 1/(2+x),\at x=2 ?
The tangent of f(x)= 1/(2+x),\at x=2 is y=-1/16 x+3/8