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

The tangent of f(x)=(3x-1)/(x^2+1),\at x=6 is y=-93/1369 x+1187/1369

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

tangent of

f (x) = \frac{3 x - 1}{x ^{2} + 1}, at x = 6

y = - \frac{93}{1369} x + \frac{1187}{1369}

Solution steps

Find the tangent point:

(6, \frac{17}{37})

Find the slope of

f (x) = \frac{3 x - 1}{x ^{2} + 1} : \frac{df ( x )}{d x} = \frac{- 3 x ^{2} + 2 x + 3}{( x ^{2} + 1 ) ^{2}}

Compute the slope of

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

x = 6 : m = - \frac{93}{1369}

Find the line with slope m=

- \frac{93}{1369}

and passing through

(6, \frac{17}{37}) : y = - \frac{93}{1369} x + \frac{1187}{1369}

y = - \frac{93}{1369} x + \frac{1187}{1369}

\int_{1}^{2} \frac{4 x ^{2} - 7 x - 12}{x ( x + 2 ) ( x - 3 )} d x

\frac{d}{d x} (x^{- \frac{3}{5}})

y = ln (x^{2} + 5)

\int \frac{1}{x x ^{2} - 16} d x

\frac{d y}{d x} = \frac{x - y ^{2}}{2 x y + y}

What is the tangent of f(x)=(3x-1)/(x^2+1),\at x=6 ?
The tangent of f(x)=(3x-1)/(x^2+1),\at x=6 is y=-93/1369 x+1187/1369