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

The tangent of f(x)=(x^2)/(x+1),\at x=2 is y= 8/9 x-4/9

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

tangent of

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

y = \frac{8}{9} x - \frac{4}{9}

Solution steps

Find the tangent point:

(2, \frac{4}{3})

Find the slope of

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

Compute the slope of

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

x = 2 : m = \frac{8}{9}

Find the line with slope m=

\frac{8}{9}

and passing through

(2, \frac{4}{3}) : y = \frac{8}{9} x - \frac{4}{9}

y = \frac{8}{9} x - \frac{4}{9}

\int 20 (e^{\frac{3}{4} x} - e^{2 x}) d x

\int_{4}^{\infty} 7 e^{- \frac{y}{2}} d y

\frac{\partial}{\partial y} (x y e^{x + y})

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

f (x) = \frac{5}{x}, (3, \frac{5}{3})

What is the tangent of f(x)=(x^2)/(x+1),\at x=2 ?
The tangent of f(x)=(x^2)/(x+1),\at x=2 is y= 8/9 x-4/9