What are the points of inflection, if any, of #f(x)=1/(x^2-4x+5) #?

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Answer 1 · 2017-03-22T23:07:12

#x=(-(-12)+-sqrt((-12)^2-4(3)(11)))/(2(3))#

Rewrite the function as:

#(x^2-4x+5)^-1#

Then differentiate:

#-(x^2-4x+5)^-2(2x-4)#

Then differentiate again. But now, we're going to have to use chain rule:

#(-(x^2-4x+5)^-2)(2)+(2x-4)(-(-2)(x^2-4x+5)^-3(2x-4))#

Rewrite:

#2(2x+4)^2/(x^2-4x+5)^3 -2/(x^2-4x+5)^2#

Use common denominators and merge the fractions:

#(2(3x^2-12x+11))/(x^2-4x+5)^3#

Inflection points may happen when #f"(x)=0# So only worry about the top part of the fraction:

#2(3x^2-12x+11)=0#

#3x^2-12x+11=0#

Use quadratic formula:

#x=(-(-12)+-sqrt((-12)^2-4(3)(11)))/(2(3))#