What are the points of inflection, if any, of #f(x)=(x-3)/(x^2-4x+5) #?

1 Answer
May 22, 2017

They are #(1,-1)#, #(4+sqrt3,(-1-sqrt3)/4)# and #(4-sqrt3,(-1+sqrt3)/4)#

Explanation:

Use the quotient rule to find #f'(x)# and #f''(x)#

You should get

#f'(x) =((1)(x^2-4x+5)-((x-3)(2x-4)))/(x^2-4x+5)#

# = (-x^2+6x-7)/(x^2-4x+5)^2#

And

#f''(x) = ((-2x+6)(x^2-4x+5)^2-(-x^2+6x-7)(2(x^2-4x+5)(2x-4)))/(x^2-4x+5)^4#

# = ((-2x+6)(x^2-4x+5)-(-x^2+6x-7)(2(2x-4)))/(x^2-4x+5)^3#

# = (2(x^3-9x^2+21x-13))/(x^2-4x+5)^3#

To see where the concavity changes analyze the sign of #f''#

The denominator is never #0# for real values of #x#.

#f''(x) = 0# where #x^3-9x^2+21x-13=0#

Observe that #x=1# is a solution (Use the Rational Zeros Theorem), and

#x^3-9x^2+21x-13 = (x-1)(x^2-8x+13)#

Find the remaining zeros by the quadratic formula.

The zeros of the numerator are #1, 4+-sqrt3#.

None of these zeros have even multiplicity, so the sign of #f''# changes at each zero. Therefore each zero is the location (#x#-value) of an inflection point.

To find the corresponding #y# values, evaluate #f(1)#, #f(4+sqrt3)#, and #f(4-sqrt3)# to get inflection points:

#(1,-1)#, #(4+sqrt3,(-1-sqrt3)/4)# and #(4-sqrt3,(-1+sqrt3)/4)#

Here is the graph of #f''(x)#