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

The tangent of f(x)= 2/(x-3),\at x=5 is y=-1/2 x+7/2

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tangent of f(x)= 2/(x-3),\at x=5

tangent of

f (x) = \frac{2}{x - 3}, at x = 5

y = - \frac{1}{2} x + \frac{7}{2}

Solution steps

Find the tangent point:

(5, 1)

Find the slope of

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

Compute the slope of

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

x = 5 : m = - \frac{1}{2}

Find the line with slope m=

- \frac{1}{2}

and passing through

(5, 1) : y = - \frac{1}{2} x + \frac{7}{2}

y = - \frac{1}{2} x + \frac{7}{2}

x \to \infty lim (\frac{( e ^{x} )}{2 x ^{2}})

\int cos (3 x) sin^{2} (3 x) d x

f (t) = \frac{4 t}{4 + t}

\int \frac{9}{1 - x ^{2}} d x

\frac{d}{d x} (ln (5 x - 4))

What is the tangent of f(x)= 2/(x-3),\at x=5 ?
The tangent of f(x)= 2/(x-3),\at x=5 is y=-1/2 x+7/2