What is the tangent of f(x)= x/(x^2+4),\at x=0 ?

The tangent of f(x)= x/(x^2+4),\at x=0 is y= 1/4 x

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tangent of f(x)= x/(x^2+4),\at x=0

tangent of

f (x) = \frac{x}{x ^{2} + 4}, at x = 0

y = \frac{1}{4} x

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

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

Compute the slope of

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

x = 0 : m = \frac{1}{4}

Find the line with slope m=

\frac{1}{4}

and passing through

(0, 0) : y = \frac{1}{4} x

y = \frac{1}{4} x

\int_{0}^{1} \frac{28}{x ^{2} - 49} d x

6 y^{^{''}} - 6 y^{^{'}} + 15 y = 0

\frac{d}{d x} (xL (x) - x)

\int \frac{7 x}{x ^{3} + 1} d x

y = (x + 7) (7 x + 2)

What is the tangent of f(x)= x/(x^2+4),\at x=0 ?
The tangent of f(x)= x/(x^2+4),\at x=0 is y= 1/4 x