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

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

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

tangent of

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

y = - \frac{1}{2} x + \frac{5}{2}

Solution steps

Find the tangent point:

(3, 1)

Find the slope of

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

Compute the slope of

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

x = 3 : m = - \frac{1}{2}

Find the line with slope m=

- \frac{1}{2}

and passing through

(3, 1) : y = - \frac{1}{2} x + \frac{5}{2}

y = - \frac{1}{2} x + \frac{5}{2}

\int 2 sin^{2} (t) d t

\frac{d}{d x} (x^{3} + 4 x)

y^{^{'}} + 2 y = 4

\int ln (3 t) d t

x \to 7 lim (\frac{x}{x ^{2} + 7})

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