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

The tangent of f(x)= x/((1+x^2)),\at x=3 is y=-2/25 x+27/50

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

tangent of

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

y = - \frac{2}{25} x + \frac{27}{50}

Solution steps

Find the tangent point:

(3, \frac{3}{10})

Find the slope of

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

Compute the slope of

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

x = 3 : m = - \frac{2}{25}

Find the line with slope m=

- \frac{2}{25}

and passing through

(3, \frac{3}{10}) : y = - \frac{2}{25} x + \frac{27}{50}

y = - \frac{2}{25} x + \frac{27}{50}

\int_{- 3}^{- 1} (x^{2} - 4 x + 8) d x

x \to - 1 lim (3^{2 x})

\frac{d}{d x} (\frac{x ^{2}}{x ^{2} - 4})

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

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

What is the tangent of f(x)= x/((1+x^2)),\at x=3 ?
The tangent of f(x)= x/((1+x^2)),\at x=3 is y=-2/25 x+27/50