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

The tangent of f(x)=(2x-1)/(x+3),\at x=3 is y= 7/36 x+1/4

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

tangent of

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

y = \frac{7}{36} x + \frac{1}{4}

Solution steps

Find the tangent point:

(3, \frac{5}{6})

Find the slope of

f (x) = \frac{2 x - 1}{x + 3} : \frac{df ( x )}{d x} = \frac{7}{( x + 3 ) ^{2}}

Compute the slope of

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

x = 3 : m = \frac{7}{36}

Find the line with slope m=

\frac{7}{36}

and passing through

(3, \frac{5}{6}) : y = \frac{7}{36} x + \frac{1}{4}

y = \frac{7}{36} x + \frac{1}{4}

\int - sin (t) d t

\frac{\partial ^{2}}{\partial x \partial y} (ln (3 x^{2} - y^{3} - 3))

\frac{d}{d x} (9 e^{- x} + e^{4 x})

\int (\frac{x ^{3}}{x ^{2} + 1}) d x

What is the tangent of f(x)=(2x-1)/(x+3),\at x=3 ?
The tangent of f(x)=(2x-1)/(x+3),\at x=3 is y= 7/36 x+1/4