What is the tangent of f(x)= 6/(x-1),(3,3) ?

The tangent of f(x)= 6/(x-1),(3,3) is y=-3/2 x+15/2

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tangent of f(x)= 6/(x-1),(3,3)

tangent of

f (x) = \frac{6}{x - 1}, (3, 3)

y = - \frac{3}{2} x + \frac{15}{2}

Solution steps

Find the slope of

f (x) = \frac{6}{x - 1} : \frac{df ( x )}{d x} = - \frac{6}{( x - 1 ) ^{2}}

Compute the slope of

f (x) = \frac{6}{x - 1}

(3, 3) : m = - \frac{3}{2}

Find the line with slope m=

- \frac{3}{2}

and passing through

(3, 3) : y = - \frac{3}{2} x + \frac{15}{2}

y = - \frac{3}{2} x + \frac{15}{2}

y = 6 x^{2} - 7 x + 4, (1, 3)

sin (3 x) cos (2 x)

\int sin^{4} (2 x) cos^{2} (2 x) d x

\int 3 sin (5 x) d x

\int \frac{1}{1 + x ^{\frac{3}{2}}} d x

What is the tangent of f(x)= 6/(x-1),(3,3) ?
The tangent of f(x)= 6/(x-1),(3,3) is y=-3/2 x+15/2