What is the equation of the line normal to $f(x)=lnx-x$ at $x=2$ ?

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Answer 1 · 2016-06-09T12:41:01

$y = log_2(2)-2+2(x-2)$

taking $f(x)=Log_e(x)-x$ the tangent space at each point is obtained
with

$m(x) = (df)/(dx) = 1/x-1$

Also the normal space direction is obtained with

$n(x) = -1/(m(x))$

In this case, at point $p_0 = {x_0,y_0} = (2,log_e(2)-2}$ we have:

$n_0(2) = -1/((-1/2)) = 2$

Finally the normal straight at point $p_0$ reads

$y = log_2(2)-2+2(x-2)$

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What is the equation of the line normal to f(x)=lnx-x f(x)=lnx-x at x=2 x=2?