What is the tangent of f(x)=(6x)/(x^2-3),(3,3) ?

The tangent of f(x)=(6x)/(x^2-3),(3,3) is y=-2x+9

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tangent of f(x)=(6x)/(x^2-3),(3,3)

tangent of

f (x) = \frac{6 x}{x ^{2} - 3}, (3, 3)

y = - 2 x + 9

Solution steps

Find the slope of

f (x) = \frac{6 x}{x ^{2} - 3} : \frac{df ( x )}{d x} = \frac{6 ( - x ^{2} - 3 )}{( x ^{2} - 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 = - 2

Find the line with slope m=

- 2

and passing through

(3, 3) : y = - 2 x + 9

y = - 2 x + 9

\frac{d y}{d x} = - 4 y + x^{3} - x

y^{^{''}} - 3 y^{^{'}} - 4 y = 3 x^{2} + 2

\int \frac{a + b x ^{6}}{7 a x + b x ^{7}} d x

4 e^{- 5 t} - 3 e^{- t}

\frac{d y}{d t} + y = sin (t)

What is the tangent of f(x)=(6x)/(x^2-3),(3,3) ?
The tangent of f(x)=(6x)/(x^2-3),(3,3) is y=-2x+9