\(\displaystyle k(t) \ = \ 2t+1; \ find \ k(-2/3)\)
\(\displaystyle k(-2/3) \ = \ 2(-2/3) +1 \ = \ (-4/3)+1 \ = \ -1/3.\)
\(\displaystyle By \ definition \ (f/g)(a) \ = \ \frac{f(a)}{g(a)}, \ g(a) \ doesn't \ equal \ zero.\)
\(\displaystyle f(a) \ = \ a^{3}-a^{2} \ and \ g(a) \ = \ 3a+3.\)
\(\displaystyle Hence, \ (f/g)(a) \ = \ \frac{f(a)}{g(a)} \ = \ \frac{a^{3}-a^{2}}{3a+3}.\)
