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

\frac{d y}{d x}, x^{\frac{2}{3}} + y^{\frac{2}{3}} = 20

y^{^{'''}} + 7 y^{^{''}} + 10 y^{^{'}} = 0

(- 4) (0.2)

3 x^{3} - 2 x^{2} - x - 5

y = \frac{4}{5} x + 1

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