What is the tangent of f(x)=(3x)/(x^2-2),\at x=1 ?

The tangent of f(x)=(3x)/(x^2-2),\at x=1 is y=-9x+6

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tangent of f(x)=(3x)/(x^2-2),\at x=1

tangent of

f (x) = \frac{3 x}{x ^{2} - 2}, at x = 1

y = - 9 x + 6

Solution steps

Find the tangent point:

(1, - 3)

Find the slope of

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

Compute the slope of

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

x = 1 : m = - 9

Find the line with slope m=

- 9

and passing through

(1, - 3) : y = - 9 x + 6

y = - 9 x + 6

\int \frac{1}{x ( x + 2 ) ( x - 3 )} d x

\int_{- \infty}^{0} \frac{1}{3 - 5 x} d x

\int \frac{x ^{3} + 21}{x ^{2} + 5 x + 6} d x

\frac{\partial}{\partial x} (a x^{b} y^{1 - b})

\frac{d y}{d x} = x^{2} e^{x^{3}}

What is the tangent of f(x)=(3x)/(x^2-2),\at x=1 ?
The tangent of f(x)=(3x)/(x^2-2),\at x=1 is y=-9x+6