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

The tangent of f(x)= 2/(x+2),\at x=1 is y=-2/9 x+8/9

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

tangent of

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

y = - \frac{2}{9} x + \frac{8}{9}

Solution steps

Find the tangent point:

(1, \frac{2}{3})

Find the slope of

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

Compute the slope of

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

x = 1 : m = - \frac{2}{9}

Find the line with slope m=

- \frac{2}{9}

and passing through

(1, \frac{2}{3}) : y = - \frac{2}{9} x + \frac{8}{9}

y = - \frac{2}{9} x + \frac{8}{9}

x \to - 1 lim (2 x^{2} - \frac{3}{x - 1})

\int \frac{b}{a ^{2} x ^{2} - c ^{2}} d x

f (x) = 5 x + y = 6

\frac{d}{d x} (6 cos^{2} (x))

What is the tangent of f(x)= 2/(x+2),\at x=1 ?
The tangent of f(x)= 2/(x+2),\at x=1 is y=-2/9 x+8/9