What is the tangent of f(x)=(2x-1)/(2x+1),(-1,3) ?

The tangent of f(x)=(2x-1)/(2x+1),(-1,3) is y=4x+7

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tangent of f(x)=(2x-1)/(2x+1),(-1,3)

tangent of

f (x) = \frac{2 x - 1}{2 x + 1}, (- 1, 3)

y = 4 x + 7

Solution steps

Find the slope of

f (x) = \frac{2 x - 1}{2 x + 1} : \frac{df ( x )}{d x} = \frac{4}{( 2 x + 1 ) ^{2}}

Compute the slope of

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

(- 1, 3) : m = 4

Find the line with slope m=

4

and passing through

(- 1, 3) : y = 4 x + 7

y = 4 x + 7

\int \frac{z ^{2}}{y} d y

\frac{( s + 1 )}{s ^{2} + s + 10}

\frac{d}{d x} (ln (\frac{x ^{3} + 4}{2 x}))

y^{^{'}} + 2 y = 2 e^{x}

\int \frac{x ^{4} + 2}{x ^{3} - x ^{2}} d x

What is the tangent of f(x)=(2x-1)/(2x+1),(-1,3) ?
The tangent of f(x)=(2x-1)/(2x+1),(-1,3) is y=4x+7