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}}

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 = - \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}

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