What is the tangent of f(x)= 1/(x-2),(1,-1) ?

The tangent of f(x)= 1/(x-2),(1,-1) is y=-x

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tangent of f(x)= 1/(x-2),(1,-1)

tangent of

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

y = - x

Solution steps

Find the slope of

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

Compute the slope of

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

(1, - 1) : m = - 1

Find the line with slope m=

- 1

and passing through

(1, - 1) : y = - x

y = - x

\int sin (7 x) sin (4 x) d x

f (x) = 6 x^{2} + 8 x - 15, at x = 2

\frac{\partial}{\partial y} (e^{- x^{2} - y^{2}})

x \to 3 lim (\frac{tan ( x - 3 )}{3 e ^{x - 3} - x})

What is the tangent of f(x)= 1/(x-2),(1,-1) ?
The tangent of f(x)= 1/(x-2),(1,-1) is y=-x