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)