What is the tangent of f(x)=(5x-7)/(x+1),\at x=0 ?

The tangent of f(x)=(5x-7)/(x+1),\at x=0 is y=12x-7

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tangent of f(x)=(5x-7)/(x+1),\at x=0

tangent of

f (x) = \frac{5 x - 7}{x + 1}, at x = 0

y = 12 x - 7

Solution steps

Find the tangent point:

(0, - 7)

Find the slope of

f (x) = \frac{5 x - 7}{x + 1} : \frac{df ( x )}{d x} = \frac{12}{( x + 1 ) ^{2}}

Compute the slope of

f (x) = \frac{5 x - 7}{x + 1}

x = 0 : m = 12

Find the line with slope m=

12

and passing through

(0, - 7) : y = 12 x - 7

y = 12 x - 7

\frac{d}{d x} (2 x + 3 x)

\frac{\partial}{\partial x} (6 e^{x y + 9})

f (x) = x^{2} - 2, at x = - 5

x \to 0 lim (2 x^{2} - 9 x + 5)

\frac{1}{( x ^{2} + 1 ) ^{4}}

What is the tangent of f(x)=(5x-7)/(x+1),\at x=0 ?
The tangent of f(x)=(5x-7)/(x+1),\at x=0 is y=12x-7