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

The tangent of f(x)= 5/(x+1),\at x=(-1)/3 is y=-45/4 x+15/4

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tangent of f(x)= 5/(x+1),\at x=(-1)/3

tangent of

f (x) = \frac{5}{x + 1}, at x = \frac{- 1}{3}

y = - \frac{45}{4} x + \frac{15}{4}

Solution steps

Find the tangent point:

(\frac{- 1}{3}, \frac{15}{2})

Find the slope of

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

Compute the slope of

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

x = \frac{- 1}{3} : m = - \frac{45}{4}

Find the line with slope m=

- \frac{45}{4}

and passing through

(\frac{- 1}{3}, \frac{15}{2}) : y = - \frac{45}{4} x + \frac{15}{4}

y = - \frac{45}{4} x + \frac{15}{4}

x \to 0 lim (\frac{0 + 5}{3 ( 0 )})

f (x) = \frac{x ^{2} + 8 x + 3}{x}

x \to \infty lim (\frac{x ^{2} + 2 x + 5}{e ^{3 x}})

\frac{d}{d x} (e^{x} cos (x) + e^{x} sin (x))

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

What is the tangent of f(x)= 5/(x+1),\at x=(-1)/3 ?
The tangent of f(x)= 5/(x+1),\at x=(-1)/3 is y=-45/4 x+15/4