Find the equation of the tangent line to the graph of fx ln

Find the equation of the tangent line to the graph of f(x) = ln(-x) at x_{0}=-e^9

Solution

given f(x) = ln(-x)

x0 = -e^9

plug x0 in f(x)

then f(x0) = ln(- (-e^9))

= ln(e^9)

= 9 ln(e)

= 9 ( lne =1)

so the point is (-e^9 , 9)

find f \'(x)

f\' (x) = -1/x

slope is nothing but f\'(x)

so slope at x0

slope = - 1/-e^9

slope = 1/e^9

so using slope point formula

y = mx +c where m =1/e^9

y = 1/e^9 x + c is passing through (-e^9 , 9)

9 = -e^9/ e^9 +c

9 = - 1 + c

c = 9+1

c=10

y = 1/e^9 x + 10 is the tangent line for f(x)