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

The tangent of f(x)= 3/(x+3),\at x=-1/3 is y=-27/64 x+63/64

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

tangent of

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

y = - \frac{27}{64} x + \frac{63}{64}

Solution steps

Find the tangent point:

(- \frac{1}{3}, \frac{9}{8})

Find the slope of

f (x) = \frac{3}{x + 3} : \frac{df ( x )}{d x} = - \frac{3}{( x + 3 ) ^{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{27}{64}

Find the line with slope m=

- \frac{27}{64}

and passing through

(- \frac{1}{3}, \frac{9}{8}) : y = - \frac{27}{64} x + \frac{63}{64}

y = - \frac{27}{64} x + \frac{63}{64}

y = x^{2}, y = 8 - 2 x

x \to 0 lim (cos (x) \cdot x)

\frac{d ^{4}}{d x ^{4}} (2 ln (x))

\frac{20}{s ( s + 30 )}

\int 2 cos (t) + sec^{2} (t) d t

What is the tangent of f(x)= 3/(x+3),\at x=-1/3 ?
The tangent of f(x)= 3/(x+3),\at x=-1/3 is y=-27/64 x+63/64