For $f (x) = \frac{2 x + 1}{x + 2}$ $f(x)=(2x+1)/(x+2)$ what is the equation of the tangent line at $x = 1$ $x=1$ ?

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Answer 1 · 2015-11-02T13:48:35

I found: $y = \frac{1}{3} x + \frac{2}{3}$ $y=1/3x+2/3$

First you need to find the slope $m$ $m$ of the tangent line. This is found deriving your function and calculating it at $x = 1$ $x=1$ :

$f'(x)=(2(x+2)-(2x+1))/(x+2)^2=(2x+4-2x-1)/(x+2)^2=3/(x+2)^2$

at $x=1$
$f'(3)=3/(1+2)^2=1/3=m$

Now we need the $y$ value of the tangence point as well; we find it setting $x=1$ into the original function:
$f(1)=(2+1)/(1+2)=3/3=1$

The equation of a line passing through $x_0=1 and y_0=1$ with slope $m=1/3$ is given as:
$y-y_0=m(x-x_0)$
$y-1=1/3(x-1)$
$y=1/3x-1/3+1$
$y=1/3x+2/3$

Graphically:
enter image source here

For f(x)=(2x+1)/(x+2) f(x)=2x+1x+2f(x)=(2x+1)/(x+2) what is the equation of the tangent line at x=1x=1x=1?