What is the tangent of f(x)= 9/(x-3),\at x=2 ?

The tangent of f(x)= 9/(x-3),\at x=2 is y=-9x+9

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tangent of f(x)= 9/(x-3),\at x=2

tangent of

f (x) = \frac{9}{x - 3}, at x = 2

y = - 9 x + 9

Solution steps

Find the tangent point:

(2, - 9)

Find the slope of

f (x) = \frac{9}{x - 3} : \frac{df ( x )}{d x} = - \frac{9}{( x - 3 ) ^{2}}

Compute the slope of

f (x) = \frac{9}{x - 3}

x = 2 : m = - 9

Find the line with slope m=

- 9

and passing through

(2, - 9) : y = - 9 x + 9

y = - 9 x + 9

y = e^{5}

\frac{d}{d x} (x^{\frac{3}{4}} - 3 x^{\frac{1}{4}})

(3.3 x + 5)^{0.37}

\frac{d}{d x} (\frac{1}{2 x ^{\frac{5}{2}}})

y^{^{'}} = (y + 9 x)^{2}

What is the tangent of f(x)= 9/(x-3),\at x=2 ?
The tangent of f(x)= 9/(x-3),\at x=2 is y=-9x+9