Is x = 11/3 an inflection point of: f(x) = (x + 1)(x - 6)^2 ?

Answer 1

Yuppers. Here's why:

f (x) = (x + 1)(x - 6)²

f '(x) = (x + 1)•2(x - 6)(1) + (x - 6)²(1)
f '(x) = (2x² - 10x - 12) + (x² - 12x + 36)
f '(x) = 3x² - 22x + 24

f ''(x) = 6x - 22

Inflection points occur when f ''(x) = 0.
6x - 22 = 0
3x - 11 = 0
3x = 11
x = 11 / 3

Answer 2

Yes, f''(11/3) = 0.
f'(x) = (x-6)^2+2*(x+1)*(x-6)
f''(x) = 6*x-22

Answer 3

Yes
f(x)=
(x+1)(x-6)(x-6)
=(x-6)(x^2-5x-6)
=(x^3-11x^2+24x+36)
1st derivative
3x^2-22x+24
2nd derivative=0
6x-22=0
so,
x=11/3