What is the tangent of f(x)=(2x)/(-x+2) ?

The tangent of f(x)=(2x)/(-x+2) is y= 4/((-a_{0)+2)^2}x-(2a_{0}^2)/((-a_{0)+2)^2}

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tangent of f(x)=(2x)/(-x+2)

tangent of

f (x) = \frac{2 x}{- x + 2}

y = \frac{4}{( - a _{0} + 2 ) ^{2}} x - \frac{2 a _{0}^{2}}{( - a _{0} + 2 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{2 a _{0}}{- a _{0} + 2})

Find the slope of

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

Compute the slope of

f (x) = \frac{2 x}{- x + 2}

x = a_{0} : m = \frac{4}{( - a _{0} + 2 ) ^{2}}

Find the line with slope m=

\frac{4}{( - a _{0} + 2 ) ^{2}}

and passing through

(a_{0}, \frac{2 a _{0}}{- a _{0} + 2}) : y = \frac{4}{( - a _{0} + 2 ) ^{2}} x - \frac{2 a _{0}^{2}}{( - a _{0} + 2 ) ^{2}}

y = \frac{4}{( - a _{0} + 2 ) ^{2}} x - \frac{2 a _{0}^{2}}{( - a _{0} + 2 ) ^{2}}

\frac{4 d u}{d t} = u^{2}

\int \frac{1}{x ( x ^{2} + 8 ) ^{2}} d x

x \to 3 - lim (a x^{2} - b x + 3)

3 t^{2} - 5 e^{3 t}

100 0^{l o g_{10} (x)}

What is the tangent of f(x)=(2x)/(-x+2) ?
The tangent of f(x)=(2x)/(-x+2) is y= 4/((-a_{0)+2)^2}x-(2a_{0}^2)/((-a_{0)+2)^2}