What is the tangent of f(x)=(8x)/(x^2+1),\at x=1 ?

The tangent of f(x)=(8x)/(x^2+1),\at x=1 is y=4

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tangent of f(x)=(8x)/(x^2+1),\at x=1

tangent of

f (x) = \frac{8 x}{x ^{2} + 1}, at x = 1

y = 4

Solution steps

Find the tangent point:

(1, 4)

Find the slope of

f (x) = \frac{8 x}{x ^{2} + 1} : \frac{df ( x )}{d x} = \frac{8 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{2}}

Compute the slope of

f (x) = \frac{8 x}{x ^{2} + 1}

x = 1 : m = 0

Find the line with slope m=

0

and passing through

(1, 4) : y = 4

y = 4

\int \frac{x ^{2}}{( x ^{3} - 5 ) ^{3}} d x

x \to - 8 lim (\frac{x + 7}{x + 8})

f (x) = ln (2 x + 1), at x = 0

u (x - 5)

f (x) = \frac{3}{x}, at x = 4

What is the tangent of f(x)=(8x)/(x^2+1),\at x=1 ?
The tangent of f(x)=(8x)/(x^2+1),\at x=1 is y=4