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

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

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

tangent of

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

y = - 2 x + 9

Solution steps

Find the tangent point:

(3, 3)

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}}

Compute the slope of

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

x = 3 : m = - 2

Find the line with slope m=

- 2

and passing through

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

y = - 2 x + 9

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

9 y^{^{''}} - 6 y^{^{'}} + 8 y = 0

\frac{d y}{d x} = 3 x^{2} (y - 1)^{3}

\frac{d}{d x} (2 (x - 2)^{2})

sin^{2} (2 t)

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