What is the tangent of f(x)= 4/(x-3),\at x=1 ?

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

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tangent of f(x)= 4/(x-3),\at x=1

tangent of

f (x) = \frac{4}{x - 3}, at x = 1

y = - x - 1

Solution steps

Find the tangent point:

(1, - 2)

Find the slope of

f (x) = \frac{4}{x - 3} : \frac{df ( x )}{d x} = - \frac{4}{( 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 = - 1

Find the line with slope m=

- 1

and passing through

(1, - 2) : y = - x - 1

y = - x - 1

\int_{1}^{4} t (1 + t) d t

\int \frac{1}{- x ^{2} - 12 x} d x

\frac{d}{d x} (x^{2} + 2 y)

\frac{d}{d x} ((arctan (x))^{\frac{1}{x}})

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

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