# Find the slope and the equation of the line tangent to f(x)= (2x-1)/(x+1) at x= 2 A) The slope of the line tangent to f(x) at x=2 is what? B) The equation of the tangent line is y=?

Find the slope and the equation of the line tangent to $f\left(x\right)=\frac{2x-1}{x+1}$ at x= 2
A) The slope of the line tangent to f(x) at x=2 is what?
B) The equation of the tangent line is y=?
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Hamnetmj
Given:
$f\left(x\right)=\frac{2x-1}{x+1}\phantom{\rule{0ex}{0ex}}\therefore {f}^{\prime }\left(x\right)=\frac{d}{dx}\left(\frac{2x-1}{x+1}\right)\phantom{\rule{0ex}{0ex}}=\frac{\left(x+1\right)\frac{d}{dx}\left(2x-1\right)-\left(2x-1\right)\frac{d}{dx}\left(x+1\right)}{\left(x+1{\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{2\left(x+1\right)-\left(2x-1\right)}{\left(x+1{\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{3}{\left(x+1{\right)}^{2}}$
$\therefore$ Slope of the tangent at x=2 is
$m={f}^{\prime }\left(2\right)=\frac{3}{\left(2+1{\right)}^{2}}=\frac{3}{\left(3{\right)}^{2}}=\frac{1}{3}$ (Answer)
Now, $f\left(2\right)=\frac{2\cdot 2-1}{2+1}=\frac{4-1}{3}=\frac{3}{3}=1$
$\therefore$ Equation of tangent at x=2 is given by