f'(x)=(x-6)(x+2)/(x-2)^2=(x^2+2x-6x-12)/(x-2)^2=(x^2-4x-12)/(x^2-4x+4)
f"(x)=[(x^2-4x-12)'(x^2-4x+4)-(x^2-4x-12)(x^2-4x+4)']/(x^2-4x+4)^2
f"(x)=[(2x-4)(x^2-4x+4)-(x^2-4x-12)(2x-4)]/(x^2-4x+4)^2
f"(x)=(2x-4)(x^2-4x+4-x^2+4x+12)/(x-2)^4
f"(x)=2(x-2)*12/(x-2)^4=24(x-2)/(x-2)^4=24/(x-2)^3
f"(x)=0
O fractie este 0 cand numaratorul este 0, deci:
24=0, ceea ce e absurd
Deci ecuatia nu are solotii si, evident, nici punct de inflexiune.
Si daca te intereseaza, functia f este convexa pe (2;inf).
x |2___________________________inf
f"|++++++++++++++++++++++++++++++++++
f | convexa convexa convexa convexa convexa
