What is the tangent of f(x)=(5x-9)/(x+1),\at x=0 ?

The tangent of f(x)=(5x-9)/(x+1),\at x=0 is y=14x-9

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tangent of f(x)=(5x-9)/(x+1),\at x=0

tangent of

f (x) = \frac{5 x - 9}{x + 1}, at x = 0

y = 14 x - 9

Solution steps

Find the tangent point:

(0, - 9)

Find the slope of

f (x) = \frac{5 x - 9}{x + 1} : \frac{df ( x )}{d x} = \frac{14}{( x + 1 ) ^{2}}

Compute the slope of

f (x) = \frac{5 x - 9}{x + 1}

x = 0 : m = 14

Find the line with slope m=

14

and passing through

(0, - 9) : y = 14 x - 9

y = 14 x - 9

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

\int 4 x 1 - x^{2} d x

\frac{d}{d x} (\frac{24}{( 1 - x ) ^{4}})

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

\frac{4}{5 x + 1}, at x = - 1

What is the tangent of f(x)=(5x-9)/(x+1),\at x=0 ?
The tangent of f(x)=(5x-9)/(x+1),\at x=0 is y=14x-9