What is the tangent of f(x)= 3/(x+1),\at x=7 ?

The tangent of f(x)= 3/(x+1),\at x=7 is y=-3/64 x+45/64

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tangent of f(x)= 3/(x+1),\at x=7

tangent of

f (x) = \frac{3}{x + 1}, at x = 7

y = - \frac{3}{64} x + \frac{45}{64}

Solution steps

Find the tangent point:

(7, \frac{3}{8})

Find the slope of

f (x) = \frac{3}{x + 1} : \frac{df ( x )}{d x} = - \frac{3}{( x + 1 ) ^{2}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - \frac{3}{64}

Find the line with slope m=

- \frac{3}{64}

and passing through

(7, \frac{3}{8}) : y = - \frac{3}{64} x + \frac{45}{64}

y = - \frac{3}{64} x + \frac{45}{64}

2 y^{^{''}} + 8 y^{^{'}} + 80 y = 60 cos (8 t)

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

n = 0 \sum \infty (x)^{n}

\int \frac{1}{t ^{4} + 16} d t

x \to - 3 + lim (\frac{x + 2}{x + 3})

What is the tangent of f(x)= 3/(x+1),\at x=7 ?
The tangent of f(x)= 3/(x+1),\at x=7 is y=-3/64 x+45/64