What is the tangent of f(x)=((3x+5))/(1+x),\at x=1 ?

The tangent of f(x)=((3x+5))/(1+x),\at x=1 is y=-1/2 x+9/2

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tangent of f(x)=((3x+5))/(1+x),\at x=1

tangent of

f (x) = \frac{( 3 x + 5 )}{1 + x}, at x = 1

y = - \frac{1}{2} x + \frac{9}{2}

Solution steps

Find the tangent point:

(1, 4)

Find the slope of

f (x) = \frac{3 x + 5}{1 + x} : \frac{df ( x )}{d x} = - \frac{2}{( 1 + x ) ^{2}}

Compute the slope of

f (x) = \frac{3 x + 5}{1 + x}

x = 1 : m = - \frac{1}{2}

Find the line with slope m=

- \frac{1}{2}

and passing through

(1, 4) : y = - \frac{1}{2} x + \frac{9}{2}

y = - \frac{1}{2} x + \frac{9}{2}

\frac{d}{d x} (3 x^{2} - 6 x - 8)

\int \frac{e ^{x}}{x - 1} d x

x \to 0 lim (ln (1 - 5 x))

x y^{^{''}} - y^{^{'}} + y^{^{'}} = 0

\frac{d}{d x} (\frac{1 + x - 4 x}{x})

What is the tangent of f(x)=((3x+5))/(1+x),\at x=1 ?
The tangent of f(x)=((3x+5))/(1+x),\at x=1 is y=-1/2 x+9/2