What is the tangent of f(x)=(4x)/(x^2+3),\at x=1 ?

The tangent of f(x)=(4x)/(x^2+3),\at x=1 is y= 1/2 x+1/2

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tangent of f(x)=(4x)/(x^2+3),\at x=1

tangent of

f (x) = \frac{4 x}{x ^{2} + 3}, at x = 1

y = \frac{1}{2} x + \frac{1}{2}

Solution steps

Find the tangent point:

(1, 1)

Find the slope of

f (x) = \frac{4 x}{x ^{2} + 3} : \frac{df ( x )}{d x} = \frac{4 ( - x ^{2} + 3 )}{( x ^{2} + 3 ) ^{2}}

Compute the slope of

f (x) = \frac{4 x}{x ^{2} + 3}

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

Find the line with slope m=

\frac{1}{2}

and passing through

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

y = \frac{1}{2} x + \frac{1}{2}

x \to 0 + lim (cos (\frac{π}{x}))

\frac{d}{d x} ((1 + x)^{\frac{1}{3}})

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

y = sin (x^{3})

\int \frac{x ^{3} - x}{x ^{3}} d x

What is the tangent of f(x)=(4x)/(x^2+3),\at x=1 ?
The tangent of f(x)=(4x)/(x^2+3),\at x=1 is y= 1/2 x+1/2