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

1 Answer
Jun 9, 2016

y = log_2(2)-2+2(x-2)y=log2(2)2+2(x2)

Explanation:

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

m(x) = (df)/(dx) = 1/x-1m(x)=dfdx=1x1

Also the normal space direction is obtained with

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

In this case, at point p_0 = {x_0,y_0} = (2,log_e(2)-2}p0={x0,y0}=(2,loge(2)2} we have:

n_0(2) = -1/((-1/2)) = 2n0(2)=1(12)=2

Finally the normal straight at point p_0p0 reads

y = log_2(2)-2+2(x-2)y=log2(2)2+2(x2)

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