What is the slope of the line normal to the tangent line of #f(x) = 1/(x^2-x+1) # at # x= 2 #?

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Answer 1 · 2016-01-08T20:16:18

slope of normal = 3

start by rewriting the function as

# f(x) = (x^2 - x + 1 )^-1 #

now using the chain rule to differentiate :

# f'(x) = - 1 (x^2 - x + 1 )^-2.d/dx (x^2 - x + 1 )#

# f'(x) = - 1 (x^2 - x + 1 )^-2(2x - 1 )#

# f'(x) =( - (2x - 1 ))/(x^2 - x + 1 )^2 #

now f'(x) = slope of tangent line and evaluating gives :

# f'(2) =( - (4 - 1 ))/(4 - 2 + 1 )^2 = -3/9 = -1/3 #

using # m_1m_2 = - 1 # for 2 lines which are perpendicular to each other.

then slope of the normal to the tangent is m = 3