What is the tangent of f(x)=(7x)/(3+x^2),\at x=2 ?

The tangent of f(x)=(7x)/(3+x^2),\at x=2 is y=-1/7 x+16/7

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tangent of f(x)=(7x)/(3+x^2),\at x=2

tangent of

f (x) = \frac{7 x}{3 + x ^{2}}, at x = 2

y = - \frac{1}{7} x + \frac{16}{7}

Solution steps

Find the tangent point:

(2, 2)

Find the slope of

f (x) = \frac{7 x}{3 + x ^{2}} : \frac{df ( x )}{d x} = \frac{7 ( 3 - x ^{2} )}{( 3 + x ^{2} ) ^{2}}

Compute the slope of

f (x) = \frac{7 x}{3 + x ^{2}}

x = 2 : m = - \frac{1}{7}

Find the line with slope m=

- \frac{1}{7}

and passing through

(2, 2) : y = - \frac{1}{7} x + \frac{16}{7}

y = - \frac{1}{7} x + \frac{16}{7}

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

y = \frac{- 6 x}{x ^{2} + 1}, (0, 0)

f (x) = 4 x^{2} - x

\int \frac{1}{9 x} d x

\int \frac{x ^{2}}{x ^{2} - 4 x + 20} d x

What is the tangent of f(x)=(7x)/(3+x^2),\at x=2 ?
The tangent of f(x)=(7x)/(3+x^2),\at x=2 is y=-1/7 x+16/7