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

Answer 1

Scarlett Allison

#y = 1 - x#

Let #y = f(x) = ln(x^2 - x + 1)/x#

#dy/dx = (x-2)/(x^2 (x^2-x+1))+(2-log(x^2-x+1))/x^2 #

Find the y coordinate at x = 1:

#y = ln(1^2 - 1 + 1)/1 = 0#

The slope, m, of the tangent line is #dy/dx# evaluated at #x = 1 #:

#m = (1-2)/(1^2 (1^2-x+1))+(2-log(1^2-1+1))/1^2 #

#m = 1#

The slope, n, of the normal line is:

#n = -1/m#

#n = -1/1#

#n = -1#

The normal line has a slope of -1 and passes through the point (1, 0). Using the point-slope form of the equation of a line, we obtain the following equation:

#y - 0 = -1(x - 1)#

#y = 1 - x#

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Answer 2

Elijah Abram

The equation of the line that is normal to f(x) = ln(x^2-x+1)/x at x = 1 is y = -1.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.