Find Inflection points of

Question

f(x)=(x^2-4)/(x^2-2x)?

Answer 1

The function f(x) = (x² - 4)/(x² - 2x)

Differentiate with respect to x.

Apply quotient rule in derivatives d/dx (u/v) = (vu' - uv')/v²

u = x² - 4 and v = x² - 2x

u' = 2x, v' = 2x - 2

f'(x) = [(x² - 2x)(2x) - (x² - 4)(2x - 2)]/(x² - 2x)²

 = [ 2x³ - 4x² - (2x³ - 2x² - 8x + 8)]/(x² - 2x)²

 = (2x³ - 4x² - 2x³ + 2x² + 8x - 8)/(x² - 2x)²

= (- 2x² + 8x - 8)/(x⁴ + 4x² - 4x³)

= - 2 (x² - 4x + 4)/x²(x² + 4 - 4x)

= - 2/x²

f'(x) = - 2 x^-2

Again differentiate with respect to x.

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

= 4x^-3

f''(x) = 4/x³

Equate second derivative = 0

4/x³ = 0

The second derivative of the function is never zero.

The second derivative is undefined at x = 0.

0 is not in the domain of the original function f(x) = (x² - 4)/(x² - 2x).

Therefore, there are no inflection points due to the fact that the original function f(x) is not defined at 0.