What is the tangent of f(x)=(x-1)/(x-2),\at x=3 ?

The tangent of f(x)=(x-1)/(x-2),\at x=3 is y=-x+5

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tangent of f(x)=(x-1)/(x-2),\at x=3

tangent of

f (x) = \frac{x - 1}{x - 2}, at x = 3

y = - x + 5

Solution steps

Find the tangent point:

(3, 2)

Find the slope of

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

Compute the slope of

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

x = 3 : m = - 1

Find the line with slope m=

- 1

and passing through

(3, 2) : y = - x + 5

y = - x + 5

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

y^{^{'}} + 2 y = 18

\int \frac{6 e ^{2 x}}{e ^{2 x} + 10 e ^{x} + 16} d x

(1 - x) e^{- x}

f (x) = 5 e^{x} - x^{2}

What is the tangent of f(x)=(x-1)/(x-2),\at x=3 ?
The tangent of f(x)=(x-1)/(x-2),\at x=3 is y=-x+5