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

The tangent of f(x)=(3x)/(x^2-6),\at x=3 is y=-5x+18

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

tangent of

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

y = - 5 x + 18

Solution steps

Find the tangent point:

(3, 3)

Find the slope of

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

Compute the slope of

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

x = 3 : m = - 5

Find the line with slope m=

- 5

and passing through

(3, 3) : y = - 5 x + 18

y = - 5 x + 18

2 y^{^{''}} - 9 y^{^{'}} - 18 y = 0, y (0) = 8, y^{^{'}} (0) = 7

\frac{d}{d x} (\frac{1 + csc ( x ^{9} )}{1 - cot ( x ^{9} )})

\int \frac{5}{x 4 - 25 ln ^{2} ( x )} d x

x (x^{4} - 1)^{6}

\int \frac{2 ln ( x )}{x} d x

What is the tangent of f(x)=(3x)/(x^2-6),\at x=3 ?
The tangent of f(x)=(3x)/(x^2-6),\at x=3 is y=-5x+18