What is the tangent of f(x)=(4x)/(x^2+1),\at x=1 ?

The tangent of f(x)=(4x)/(x^2+1),\at x=1 is y=2

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tangent of f(x)=(4x)/(x^2+1),\at x=1

tangent of

f (x) = \frac{4 x}{x ^{2} + 1}, at x = 1

y = 2

Solution steps

Find the tangent point:

(1, 2)

Find the slope of

f (x) = \frac{4 x}{x ^{2} + 1} : \frac{df ( x )}{d x} = \frac{4 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{2}}

Compute the slope of

f (x) = \frac{4 x}{x ^{2} + 1}

x = 1 : m = 0

Find the line with slope m=

0

and passing through

(1, 2) : y = 2

y = 2

\int \frac{1}{x x ^{2} - 5} d x

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

\int - 14 x d x

\int \frac{1}{3} e^{3 x + 1} d x

y = sin (4 x), (0.1, 0.4)

What is the tangent of f(x)=(4x)/(x^2+1),\at x=1 ?
The tangent of f(x)=(4x)/(x^2+1),\at x=1 is y=2