Let f(x)=x^3 for x≤1 and f(x)=ax^2+bx+c for x>1. If f, f' and f'' are all continuous at x=1, then abc equals

At $x=1$ the left side has $f=1$, $f'=3x^2=3$ and $f''=6x=6$. Continuity of $f$ requires $a+b+c=1$. For $f'$: $2a+b=3$. For $f''$: $2a=6$. Hence $a=3$, then $b=3-6=-3$, and $c=1-a-b=1-3+3=1$. Therefore $abc=3\cdot(-3)\cdot1=-9$. Note that matching only $f$ and $f'$ would leave one parameter free; the extra condition on $f''$ is what makes the answer unique. The pattern to remember is that continuity of the $k$-th derivative at the joining point gives one linear equation per derivative, so three unknown coefficients need three matching conditions. As a final check, the right piece $3x^2-3x+1$ equals $1$, $3$ and $6$ for the function and its first two derivatives at $x=1$.